366 DiDerential Equations [Ch. 9

3107*. For a rectilinear homogeneous rod whose axis coincideswith the x-axis, the temperature u = u (x, t) in a cross-section withabscissa x at time t, in the absence of sources of heat, satisfiesthe equation of heat conduction

au ! a2uat =a ax! ,

where a is a constant. Determine the tenlperature distributionfor any tinle t in a rod of length 100 cm if we know the initialtenlperature distribution

U (x t 0) = 0.01 x (IOO-x).

Chapter X

APPROXIMATE CALCULATIONS

Sec. 1. Operations on Approximate Numbers

to. Absolute error. The ab~olute error of an approximate number a whichreplaces the exact number A is the absolute value of the difference bet\\'eenthem. The number ti, which satisfies the inequality

IA-al~L\, (1)

is called the limIting absolute error. The exact nUlnber A is located withinthe litnits a-&~A~a+& or, more briefly, A=a 6.

2. Relative error. By the relative error of an approximate nUlnber areplacing an exact number A (A > 0) we understand the ratio of the absoluteerror of the number a to the exact nurnber A. The number ~, which satisfiesthe ineq uality

I A-at ~A 1Iilrl::a, (2)

is called the limiting relative error of the approximate number a. Since in

actual practice A =::::a, we often take the number 6=~ for the lilnitinga

relative error.3. Number of correct decimals. We say that a positive approximate

number a written in the fornl of a decimal expansion has n correct decimalplaces in a narrow sense if the absolute error of this number does not exceedone half unit of the nth decinlal place. In this case, when n > I we cantake, for the limIting relative error, the number

6=2~(I~r-1 where k is the first significant digit of the number a. And conversely, if it

1 ( 1 ' n-lis known that 6E;;; 2 (k + I) 10 ) . then the number a has n correct deci malplaces in the narrow meaning of the word. In particular, the number a

definitely has n correct decimals in the narrow meaning if f> E;;; ~ (/0) n .If the absolute error of an approximate nUlnber a does not exceed a

unit of the last decimal place (such, for example, are numbers resultingfrom measurements made to a definite accuracy). then it is said that alldecimal places of this approximate number are correct in a broad sense. Ifthere is a larger number of significant digits in the approxinlate number,the latter (if it is the final result of calculations) is ordinarily rounded offso that all the remaining digits are correct in the narrow or broad sense.

368 Approximate Calculations [Ch. 10

Hencetorth, we shall assume that all digits in the initial data arecorrect (if not otherwise stated) in the narrow sense. The results of inter-mediate calculations may contain one or two reserve digits.

We note that the exanlples of this sectiCJn are, as a rule, the results offinal calculataons, and for this reason the answers to them are given asapproximate numbers with only correct drcimals.

4. Addition and subtraction of approximate numbers. The limiting ab-solute error ot an algebraic sum of sevE'ral numbers is equal to the sum ofthe limiting absolute errors of these numbers. Therefore, in order to have,in the sum of a ~mal1 number of approximate numbers (all decimal placesof which are correct), only correct digits (at least in the broad sense), allsummands should be put into the form of that summand which has thesmallest number of decimal places, and in each summand a reserve digitshould be retained. Then add the resulting numbers as exact numbers, andround off the sum by one decimal place

If we have to add approximate numbers that have not been rounded off,they should be rounrled off and one or two reserve digits ~houl d be retained.Then be guided by the foregoing rule of addition while retaining the appro-priate extra digits in the sunl up to the end of the calculations.

Example 1. 215.21 +14.182+21.4==215.2l1)+14.1(8)+21 4=250.8.The relative error of a sunl of poc:;itive terms bes between the least and

greatest relative errors of these terms.The relative error of a d,n'erence is not amenable to simple counting.

Particularly unfavourable in this sense is the ditference of two close nunlbers.Example 2. In subtracting the a~proximate numbers 6 135 and 6.131 to

four correct decimal places, we get the dIfference 004. The limiting relative~ 0.00\ + ~ 0/'01 I

error is B 0.004 =4=0.25. Hence, not one of the decimalsof the difference is correct. Therefore, it is always advisable to avoidsubtracting close approximate numbers and to trttnsform the given expression,if need be, so that this undesir~!-Ie operation is omitted.

5. Multiplication and division of approximate numbers. The limitingrelative error of a product and a quotient of approximate numhers is equal10 the sum of the linliting relative errors of these numbers Proceeding frolnthis and applying the rule for the number of correct decilTaals (3), we retainIn the answer only a definite nunlber of decimals

Example 3. The product of the approximate numbers 25.34.12 = 104.236.Assuming that all dEcimals of the factors are correct, we find that the

limiting relative error of the product is

1 I6=2.20.01 + 4.2.01 ~0.003.

Whence the number of correct decimals of the product is three and theresult, if it is final, should be written as follows: 25.34 12 = 104, or morecorrectly, ~5 34.12= 104 2 (J.3.

6. Powers and roots of approximate numbers. The limitIng relative errorof the mth power of an approxi mate number a is equal to the m-fold linlitingrelative error of this number

The limiting relative error of the mth root of an approximate number a

is the ..!.- th part of the limiting relative error of the number a.m

7. Calculating the error of the result of various operations on approxi-mate numbers. If lill., ... flail are the limiting absolute errors of the appro-

Sec. I) Operations on Approximate Numbers 369

xirnate numbers all _ an' then the limiting absolute error ~S of the result

S === f (at~ . . . an) .may be evaluated approximately from the formula

~S=I:~ I~al++I:!n IM nThe limiting relative error S is then equal to

6S = ~; I= I:1.1 f;; + .+I:!nIt'tr =dln'l Idln'l= da

lL\a l + .+ dan tian

Example 4. Evaluate S = In (10.3+ V4_4); the approximate numbers10.3 and 4.4 are correct to one decimal place.

Solution. Let us first compute the limiting absolute error L\S in the

general form: 8=ln (a+ vb), ~S=a+~b(~a +~ ;bb)' We have~a=M~ ;0; V4."4 = 2.0976... ; we leave 2.1, since the relative error of

the approximate number 4.4 is equal to ~} ~= ;0; the absolute error

is then equal to ~ 2 8~) = ~; we can be sure of the first decimal place. Hence,

1 ( 1 1 1) 1 ( I) 13L\S=IO.3+2.1 20+2--20.2.1 =12.4.20 l+U =2604=::::0.005.

Thus, hvo decimal places will be correct.Now let liS do the calculations with one reserve decinlal:

log (10.3 + y44) ~ log 12 4 -== 1.093, In (10 3+ yU)~ 1.0932.303 = 2.517.And we ~('t the answer: 2 52

8. Establishing admissible errors of approximate numbers for a givenerror in the result of operations on them. Arplying the formulas of 7 forthe quantities L\S or f,S given us and conSIdering all particular differentials

Ia~ I~ak or the quantities Ia~ I~,ai equal, we calculate the admissibleab~olute errors L\a" ... , L\an , " of the approximate numbers at ... an' that enter into the operations (the principle of equal effects).

It should be pointed out that sOlnetimes when calculating the adlnissibleerrore;; of the argul11ents of a function it is not advantageous to use theprinciple of equal effects. since the latter nlay make delllands that arepractically unfulfilable In these cases it is advisable to nlake a reasonableredistribution of errors (if thi~ is possible) so that the overall total error doesnot exceed a specified quantity. Thus, strictly speaking, the problem thusposed is indeterm inate.

Example 5. The volume of a "cylindrical segment", that is,''a ~olid cutoff a circular cylinder by a plane passing through the diameter of the base(equal to 2R) at an angle a to the base, is computed from thp formula

V = ; R' tan a. To what degree of accuracy should we measure the radius

370 Approximate Calculations [Ch. 10

R:::::: 60 cm and the angle of inclination a so that the volume of the cylindricalsegment is found to an accuracy up to 1%?

Solution. If l\V, l\R and L\a are the limiting absol ute errors of thequantities V, R and a, then the limiting relative error of the volume V thatwe are calculating is

3L\R 2L\a 16=R+sin2a~100

3L\R 1 2L\a 1We assume R ~ 200 and sin 2a.~ 200 Whence

R 60 cml\R E;;; 600 ~ 600 = 1 mIn;

A sin 2a 1 d 9'ua ~ 400

Sec. 1] Operations on Approximate Numbers 371

3115. The sides of a rectangle are 4.02 and 4.96 m (accurateto 1 cm). Compute the area of the rectangle.

3116. Find the quotient of the approximate numbers, whichare correct to the indicated decimals:

a) 5.684 : 5.032; b) 0.144 : 1.2; c) 216:4.

3117. The legs of a right triangle are 12.10 cm and 25.21 cm(accurate to 0.01 cm). Compute the tangent of the angle oppositethe first leg.

3118. Compute the indicated powers of the approximatenumbers (the bases are correct to the indicated decimals):

a) 0.41582 ; b) 65.21 ; c) 1.52

3119. The side of a square is 45.3 cm (accurate to I mm).Find the area.

3120. Compute the values of the roots (the radicands arecorrect to the indicated decimals):

a) Y2.715; b) V65.2; c) J,/81.1.

3121. The radii of the bases and the generatrix of a truncatedcone are R=23.64 cmO.OI cm; r-=17.31 clnO.OI cm; 1== 10.21 em + 0.01 cm; 1t == 3.14. Use these data to compute thetotal surface of the truncated cone. Evaluate the absolute andrelative errors of the result.

3122. The hypotenuse of a right triangle is 15.4 cm + 0.1 cm;one of the legs is 6.8 cm 0.1 cm. To \\yhat degree of accuracycan we determine the second leg and the adjacent acute angle?Find their values.

3123. Calculate the specific \veight of aluminiulTI if an alumin-ium cylinder of diameter 2 cm and altitude II em weigh593.4 gmt The relative error in measuring the lengths is 0.01,while the relative error in weighing is 0.001.

3124. Compute the current if the electromotive force is equalto 221 volts 1 volt and the resistance is 809 ohms I ohm.

3125. The period of oscillation of a pendulum of length l isequal to VT

T=21t - ,g

where g is the acceleration of gravity. To what degree of accuracydo we have to measure the length of the pendulum, whose periodis close to 2 sec, in order to obtain its oscillation period with arelative error of 0.5%? How accurate must the numbers 1t and gbe taken?

3126. It is required to nleasure, to within 1%, the lateralsurface of a truncated cone whose bas~ radii are 2 m and 1 m,and the generatrix is 5 m (approximately). To what degree of

872 Approximate Calculations [Ch, 10

accuracy do we have to measure the radii and the generatrix andto how many decimal places do we have to take the number n?

3127. To determine Young's tnodulus for the bending of arod of rectangular cross-section we use the formula

I liP=4 d3bs'

where I is the rod length, band d are the basis and altitude ofthe cross-section of the rod, s is the sag, and P the load. Towha t degree of accuracy do we have to measure the length landthe sag s so that the error E should not exceed 5.50A) , providedthat the load P is known to 0.1 ~u, and the quantities d and bare known to an accuracy of 1%, I ~ 50 em, S~ 2.5 cm?

Sec. 2. Interpolation qf Functionsto. Newton's interpolation formula. Let Xo, Xl' , Xn be the tabular val-

ues of an argument, the difference of which h=~xj (L\Xj=Xj+l-xi; i=O,I,.. , n - 1) is constan t (table Inter val) and YOt Yl' ., YtI art' the correspon d-in~ values of the function y Then the valup of the function y for an inter-mediate value of the argument x is approximately given by Newton's inter-polation formula

Y=Yo+q.,1yo+ q (q2~ I) ,12yO + ... +q (q-I). n~q-n+ I) ,1nyo (1)x-Xowhere q = -h- and ~Yo= YI- Yo, L\2yo = L\y.- !lyo, are successIve fini te

dU'erences of the furction y. \\ hen x=Xj (t=O, 1, , n), the ro1ynomial(1) takes on, accorciingly, the tabular values Yi (l =0, 1, .. , n). As partic-ular cases of l'\ewtc,n's fOrlllula we obtain: for n= 1, linear Interpolation;for n ~ 2, qupdratlc Interpolation. To simplify the use of Newton's fonnula,it IS advisable first to set up a table of finite differences.

If Y=f (x) is a polynomial of degree n, then

L\nYi = const and ~n+IYi=-O

and, hence, formula (1) i~ exactIn tile general case, if f (x) ha') a continuous derivative f (n+J) (x\ on the

interval la, b), which includes the points xo, Xl' , xn and x, then the errorof formula (1) is

R ()- "n.... q(q-l) ... (q-i+l) ~' _n X - Y- k.. If Yo-

;=0

(2)

whE're ~ is some intermediate value between xi (I =0, 1, ... , n) and x. Forpractlcal use, the {ollowing approximate formula is more convenient:

Sec. 2) Interpolation 0/ Functions 373

If the number n may be any number, then it is best (0 choose it so thatthe difference ~n+ lyO::::: 0 w,thin the "mits of the given accuracy; in otherwords. the difterences 6.'lyo should be constant to wi thin the given places ofdecimals

Example 1. Find sIn 26"15' using the tabular data sin 260 =0,43837.sin 27 ~ 0.45399, sin 28 -:- 0.46947.

Solution. We set up the table

X, I Y, I AI/, I A'II.o 26 0 42837 I 1562 I --141 27(1 0 45399 15482 28 0 46947

261fi'-26 1Here, 11=60', q= 60' 4 .

Appl ying formula (1) and uSing the first horizontal line of the table, wehave

.!- ('!--l )sin 2615' =0.43837+ f 0.01562+ 4 ~, '. (-0.00014)=0.44229.Let us evaluate the error R2 USing formula (2) and taking into account

that if y = Sill x, then I y,nl I~ 1, we will have:

f(f- I )({-2)(3t)3 7 I 1 _IIR21~ 31 180 =12d5rl.333~410.

Thus, all the decimals of sin 2015' are corr~ct.Using Newton's formula, it is alsc pOSSible, from a ~iven intermediate

value of the function y. to find the correspolldlng value of the argument x(inverse interpolation). To do this. first deterrnlne the corresponding value qby the method of successIve approx,nlatlon, putting

q'O) =Y-Yotiyo

and(I) ((I) 1) A2

q(l+l)==q(O)_L:L-=-. -lh._21 I!J.Yo

(I = 0, 1. 2. . .. ).

Here, for q we take the common value (to the given accuracyl) of two suc-cessive approximations qC1n) '='lcm+l). Whence x=xo+q.h.

Exanlple 2. USln!! the table

I II=- ,lnl1 J2 2 I 4.4572 4 5.4662 6 6.695

1.0091 229

0.220

approximate the root of the equation Sinh x = 5.

'874 Approximate Calculations (Ch 10

Solution. Taking Yo = 4.457, we have

(0)_ 5-4.457 0.543 -0 538.q - 1.009 =1.009- ,

. (I) _ (0) + q(O) (l_q(O A2yO =0 538+0.538.0.462 0.220 =q -q 2 ~Yo' 2 1.009

= 0.538+0 .027 = 0.565;qIZ)=0.538+ 0.565;0.435. ~:~:=0.538+0.027=0.565.

We can thus takex= 2.2+0.5650.2 =2.2 +0.113=2.313.

2. Lagrange's interpolation formula. In the general case, a polynomial ofdegree n, which for x=xi takes on given values Yi (i =0, I, ... , n), is givenby the Lagrange interpolation formula

JJ = (X-Xl~ (X-Xz\ . .. \X-Xn) ) Yo + X-Xo~ ~X-X2)) .(~-Xn) )YI +...(XO-x1 (XO-x1 Xo-Xn X1-XO XI -X2 X1-Xn

+ (x-xo) (X-Xl)" .(X-Xk_l) (X-Xk+l)" . (X-Xn) + (Xk-Xo) (Xk-Xl)" (Xk- Xk-I)(Xk- Xk+l)' .. (Xk-Xn) Yk

+ (X-Xo) (X-Xl)" .(X-Xn - 1)... (Xn-Xo) (Xn-XJ) (Xn-Xn - l ) Yn

3128. Given a tahie of the values of x and y:

x I 2 I 3 I 4 5 6y I 3 10 I 15 I 12 9 5

Set up a tabl~ of the finite differences of the function y.3129. Set up a table of differences of the function y=xs _

-5x2 +x,-l for the values x= 1, 3, 5, 7, 9, II. Make sure thatall the finite differences of order 3 are equal.

3130*. Utilizing the constancy of fourth-order differences, setup a ta ble of differences of the function y = x4 - IOx l +2x -t- 3xfor integral values of x lying in the range I~ x~ 10.

3131. Given the tablelog 1= 0.000,log 2 = 0.301,log 3 = 0.477,log 4 = 0.602,log 5 = 0.699.

Use 1inear interpolation to compute the numbers: log 1.7, log 2.5,log3.i, and log 4.6.

Sec. 2] Interpolation of Functions 375

3132. Given the tablesin 10=0.1736, sin 13=0.2250,sin 11=0.1908, sin 14=0.2419,sin 12 = 0.2079, sin 15 = 0.2588.

Fill in the table by computing (with Newton's formula. for n = 2)the values of the sine every half degree.

3133. Form Newton's interpolation polynomial for a functionrepresented by the table

40 I 853134*. Form Newton's interpolation polynomial for a function'

represented by the table

x I 2 I 4 I 6 I 8 10y I 3 I 11 I 27 I 50 83

Find y for x=5.5. For what x will y=20?3135. A function is given by the table

x -2 I I 2 I 4Y 25 -8 1-151 -23

Form Lagrange's interpolation polynomial and find the value ofy for x=O.

3136. Experiment has yielded the contraction of a spring (x 111IU)as a function of the load (P kg) carried by the spring:

x I 5 I 10 15 20 25 30 35 40P I 49 I 105 172 253 352 473 619 793

Find the load that yields a contraction of the spring by 14 rom..

376 Approximate Calculations [Ch. 10

3137. Given a table of the quantities x and y

x 0 131415

y -3 I 25 1 129 I381\A>mpute the values of y for x = 0.5 and for x = 2: a) by meansof linear interpolation; b) by Lagrange's formula.

Sec. 3. Computing the Real Roots of Equations

10 Establishing initial approximations of roots. The approximation of theroots of a given equation

f (x) =0 (1 )

consists of two stages: 1) separating the roots, that is. establishing the inter-vals (as small as possible) within which lies one and only one root of equa-tion (1); 2) computing the roots to a given degree of accuracy

If a function I (..\) is defined and continuous on an interval [a. b] andf(a).' (b) < 0. then on [a. b) thrre is at least one root ~ of equaticn (1).This root will definitely be the only one if I' (x) > 0 or f' (x) < 0 whena

Sec. 3] Computing the Real Roots of Equations 377

(3)

(5)

x=x-Af (x),

where the number A:. 0 is chosen so that the function d!! [X-AI (x = I-Af' (x).t

en' we can make use of the formulaP:-c 1~lf(cn)1~ n J..t'

where J..t = min f I' (x) Ia~.t 0, then in fornlula'; (3) and (3') we should put xo=b.4. Iterative Inethod. Let the given equation be reduced to the form

x==q>(x), (4)

where' (Xl)' . , xu=q> (X"_l)'

If a

.378 Approximate Calculations [Ch. 10

should be small in absolute value in the neighbourhood of the point Xo (forexample, we can put I-At' (xo) =0].

Example 1. Reduce the equation 2x-ln x-4 =0 to the form (4) for theinitial approximation to the root Xo= 2.5..

Solution. Here, f(x}=2x-lnx-4; f'(X)=2-1.. We write the equiva-x

lent equation X=X-A (2x-ln x-4) and take 0.5 as one of the suitableval ues of A; this number is close to the root of the equation

,II-A ( 2_..!-) I_ =0, that is, close to 1..!-6:::::: 0.6.I x x-2.5 .

The initial equation is reduced to the formx=x-0.5 (2x-lnx-4)

orI

x=2+2"lnx.

Example 2. Compute, to two decimal places, the roof; of the preceedingequation that lies bet\veen 2 and 3.

Computing the root by thJ; iterative method. We make use of the resultof Example 1, putting Xo= 2.5. We carry out the calculations using formulas(5) with one reserve decimal.

Ix t =2+2" In 2.5::::::2.458,

Ix2=2 +"2 In 2.458:::::: 2.450,

Ixa=2+ 2 In 2.450:::::: 2.448,

Ix4 =2 +2 In 2.448 ~ 2.448.

And so ~ ~ 2 45 (we can stop here since the third decimal place has'become fixed)

Let us nQw \evaluate the error. Here,

lp(X)=2+{lnx and lp'(X)=21x.

Considering that all approximations to xn lie in the interval [2.4, 2.5], weget

f=max I lp' (x) '=2.~.4=0.21.

Hence, the limiting absolute error in the approximation to x. is, by virtueof the remark made above,

~= 1~~~1 0.0012::::: 0.001.

Thus, the exact root ~ of the equation lies within the limits

2 447 < ~ < 2.449;we can take '~2.45, and all the decimals of this apprOXimate number willbe correct in the narrow sense.

Sec. 8) Computing the Real Roots of Equations 379'

Calculating the root by Newton's method. Here,

f(x)=2x-lnx-4, f' (X)=2-~,

On the interval 2~ x~ 3 we have: l' (x) > 0 and r (xl> 0; f (2) f (3) < 0;f (3) 1" (3) > O. Hence, the conditions ()f 3 for Xo=3 are fulfilled.

We take

(1 )-1

Q= 2-3 =0.6.

We carry out the calculations using formulas (3') with two reserve decimals:

x1 =3-0.6(2.3-ln3-4)=24592;XI =2.4592-0.6 (22 4592-ln 2 4592 -4) = 2 4481;x, = 2. 4481-0.6 (2 2. 4481-ln 2.4481-4) = 2. 4477;x4 =2.4477 -0.6 (2 2 4477 - In 2.4477-4) =2 4475.

(6){f (x, y) =0,

q> (x, y) =0,

At this stage we stop the calculations, since _the third decimal placedoes not chan~e any more. The answer is: the root. ;=2.45. We olnit theevaluation of the error.

5. The case of a system of two equations. Let it be required to calcu-late the real roots of a system of two equations in two unknowns (to a glvendegree of accuracy):

and let there be an initial approxinlation to one of the solutions (~, l)) ofthis system x = Xo' Y= Yo'

This initial approximation may be obtained, for example, graphically.by plotting (in the sallle Cartesian Loordinate systenl) the curves f (x, y) =0and q> (XI Y) ::::::0 and by determining the coordinates of the points of inter-section of these curves.

a) Newton's method. Let us suppose that the functional determinant

1=a(I, qa (x, y)

does not vanish near the initial approximation x=xo' y=Yo' Then by New-ton's method the first approxitnate solution to the system (6) has the formXI = Xo+uo' Yl = Yo + ~o' where ao' ~o are the solu tion of the syst~ln of two.linear equa lions

{

I (xo' Yo) +aof~ (xo Yo) +~of~ (xo, Yo) = O.q> (xo, Yo) +o.oq>: (xo, Yo) + ~oq>~ (xo, Yo) = O.

The second approximation is obtained in the very same way:

X2=X1 +Ul , Y2=Yl +Pl'where a p ~I are the solution of the systeln of linear equations

{f (Xl' YI) +alf~ (Xl' YI) + P.f~ (Xl' YI) =0,q> (x I' !II) +U Icp~ (xI' YI) +~ Icp;, (Xl' YI) = O.

Similarly \ve obtain the third and succeeding approximations.

380 Approximate Calculat tons (Ch 10

(8)

b) Iterative method. We can also apply the Iterative method to solvingthe system o( equations (6), by transforming this 'iystem to an equivalent one

{x= F (x, y), (7)y=cD (x, y)

and assuming that

IF~(x.y)I+I

Sec. 3] Computing the Real Roots of Equations 381

Solution. Here, I (x, y) = X2 +y!-l, q> (x, y) =xl - y; I~ (xo, Yo) = 1.6,"" (xo, Yo) = 1.1; q>~ (xo' Yo) = 1.92, q>~ (xo' Yo) = - 1. .

Write down the system (that is equivalent to the initial one)

{a (x

2 +y2_1) -t- ~ (x3 _y) =0, (I a, ~ 1#= 0)V(x2+y2-1)+~(xl_y)=O y, l>

in the formx=x+a(xl+yl-l)+~ (xl_y),y = y +V (x2 + y2_1) + ~ (xl - y).

For suitable numerical values of a, ~, y and l> choose the solution of thesystem of equations

{

1+1.6a+l.92~=0,

1.1a-~=0,

1.6y -t- 1.92~ = 0,l+l.ly-~==O;

l. e., we put a ~-O.3, ~ ::::::-0.3, V:::::::-0.5, 6~ 0.4.Then the system of equations

{x=x-O.3(x2+y2-1)-0.3(xl-y),y == y -0.5 (x2 + y2_1) +0 4 (xl_y),

which is equivalent to the initial system, has the fornl (7); and in a suffi-ciently sInall neIghbourhood of the pOInt (xo, Yo) condition (8) will be fulfilled.

Isolate the real roots of the equations by trial and error, andby means of the rule of proportional parts compute thenl to twodecinlal places.

3138. x3 - x -I 1===0.3139. x4-~-05x-I.55=0.3140. x l -4x --I == O.Proceeding fronl the graphically found initial approximations,

use Ne\vton's rnethod to compute the real roots of the equationsto two decimal places:

3141. x l -2x-5=O. 3143.2x ==4x.13142.2x-Inx-4=0. 3144.logx=-.x

Utilizing the graphically found initial approximations, use theiterative method to C01l1pute the real roots of the equations totwo decirna I places:

3145. xl -5x--t 0.1 =0. 3147. x'-x-2=O.3146. 4x == cos x.Find graphically the initial approximations and compute the

real roots of the equations and systerns to two decinlals:3148. x'-3x+ 1 =0. 3151. xlnx-14=0.3149. xl -2x2 +3x-5 = O. 3152. Xl +3x-0.5 = O.3150. x4 +Xl - 2x-2 = O. 3153. 4x-7sin x = o.

382 Approximate Calculations [Ch. 10

3157. { x2

+y-4=O,y-Iogx-l =0.

3154. XX +2x-6 = O.3155. eX +e- 1x-4 = O.3156. {xl + y2 -1=O,

xl_y=O.3158. Compute to three decimals the smallest positive root of

the equation tan x = x.3159. Compute the roots of the equation x tanh x = 1 to four

decimal places.

Sec. 4. Numerical Integration of Functions

1. Trapezoidal formula. For the approximate evaluation of the integralb

) f (x) dxa

[I (x) is a function continuous on [a, b]] we divide the interval of integratiol1[a, b] into n equal parts and choose the interval of calculations h = b-a .

11Let xi=xo+ih (xo=a, xn=b, i=O, 1, 2, ... , n) be the abscissas of the par-ti tion points, and let Yi = f (xi) be the corresponding values of the integrandY= f (x). Then the trapezoidal formula yields

b

Sf(x)dx=::h (Yo~Yn+YI+Y.+"'+Yn_l) (1)a

with an absolute error ofh2

Rn ~ 12 (b-a).M 1,

where M2=maxlr(x)1 when a

Sec. 4) Numerical Integration 01 Functions 383

holds with an absolute error ofh4

Rn ~ 180 (b-a) M" (4)

where M4 =max It lV (x) I when a~x

384 Approximate Calculations [Ch. 10

1

3162. Using the Simpson formula, calculate 5:~Xl to fOUfo

decimal places, taking n = 10. Establish the upper limit d of abso-lute error, using the error formula given in the text.

Calculate the following definite integrals to two decimals:1 2

3163. 5 dx 3168. 5SI~X dx.l+x0 0

I n

3164. 5 dx 3169. 5SI: x dx.1+x2

0 0

I 2

3165. 5 dx 3170. 5co;x dx.1+x,0 In

I

~ xlogxdx.2

3166. 3171. 5cos x d1 l+x x.

I 0

5IO~x dy. 13167. 3172. ~ e- X2 dx.I 0

3173. Evaluate to two decimal places the improper integralU)

51~XX2 by applying the substitution x = +. Verifl the calculationsby applying Simps~~'s formula to the integral 51~\2' where bis chosen liO that 5l~XI

Sec. 5] Numerical Integration of Ordinary DiOerential Equations 385

The solution y (x) of (1), which satisfies the given initial condition, can,generally speaking, be represented in the form

y (x) = lim Yi (x)1-+ r:D

(2)

where the suCCeStHVe approximations Yi (x) are determined from the fonn ulas

Yo (x) = Yo,x

YI (x) =Yo+ ~ f (x, YI_I (x)) dxXo

(i = 0, 1, 2, ... ).

If the right side f (x, y) is defined and continuous in the neighbo~lrhood

R{lx-xoJ~a, IY-Yo'~b}

and sa t isfies, in this neighbourhood, the Lipschitz cond itlon

(L. is constant), then the process of successive approximation (2) definitelyconverges in the interval

Ix-xol~h,

where h = m~n ( a, ~) and M= m;x I f (x, y) I. And the error here is

_ n I X-Xo In+lRn-,y(x)-Yn(x)I~/v'L (n+I)1 '

IfIx-xo ' ~ h.

The method of successive approximation (Picard's method) is also appli-cable, with slight nl0difications, to normal systems of differential equations.Differential equations of higher orders rnay be written in the fonn of systemsof differential equations.

2. The Runge-Kutta method. Let it be required, on a given IntervalXo~ x~ X, to find the solution Y (x) of (1) to a specified degree of accuracy 8.

To do this, we choose the interval of calculations h=X-xo by dividingn

the interval [xo, Xl into n equal parts so that h4 < 8. The partition pointsxi are determined from the formula

xl=Xo+ih (t =0, 1, 2, ... , n).

By the Runge-Kutta method, the corresponding values Yi=Y (x;) of the desiredfunction are successively computed fron] the formulas

Yi+1 =Yi+ ~Yi.

~Y'-.!. (k(i) +2k(i)+2k(t)+k(i'-6 1 I ..,

13-1900

38.::.6 .-,.;A:..:.!:..p!:....pr:....:o:.....x--.im_at_e_C_a_lc_u_la_t_io_n_s [C_"'h_._10

where

(3)

(6)

To check the correct choice of the interval h it is advisable to verifythe quantity

I

k(i)-k(t) Ie_ 2 a- k~t)-k~t)

The fracti'Jn e should amount to a few hundredths, otherwise h has to bereduced.

The Runge-Kutta method is accurate to the order of hI. A rough estiIn ateof the error of the Runge-Kutta method on the given interval [xo, Xl Illaybe obtained by proceeding from the Runge principle:

R = IY2rn--Ym 115 'where n = 2rn, Y2tn and Ym are the results of calculations using the schenle (3)\vith interval h and interval 2h.

The Runge-Kutta method is also applicable for solving SYStClllS of dlffe~rential equations

y'=f(x, y, z), z'-=cp(x, y, z) (4)

with given initial conditions Y==Yo, z=zo when x==xo.3. Milne's method. To solve (1) by the Mtlne nzethod, subject to the

initial condi\ions Y=Yo when x=xo, we in SOlne way find the successIvevalues

YI = Y (XI)' Y2 = Y (x 2), Y, == Y (xa)of the desired function y (x) [for instance, one can expand the solution Y (X)in a series (Ch. IX, Sec. 17) or find these values by the lTlethod of successi veapproximation, or by using the Runge-Kutta Inethod, and so forth]. The ap-proximations Yi and Yi for the following values of Yi (i =4, 5, ... , n) aresuccessively found frool the formulas

Yl=Yi-.+ ~h (2/; __ -/;__ +21;_.), }&:: h _ (5)Yi=Yi-2+ 3" (Ii +4Ii-l + Ii-I)'

"'here fi= f (xi, Yi) and fl= f (xi, Yl). To check we calculate the quantity1 1- =:7 1ei=29 YI-Yi

Sec..5] Numerical Integration of Ordinary Differential Equations 387

If 8i does not exceed the unit of the last decimal 10- m retained in theanswer for Y (x). then for Yi we take !Ii and calculate the next value Yi+t,repeating the pr8cess. But If 8i > 10-tn, then one has to start from the be-g1nning and reduce the interval of calculations. The Illagnitude of the initialinterval is detennined approXinlately fronl the inequality hoi < lo-m.

For the case of a solution of the system (4), the Milne formulas arewritten separately for the functions y (x) and z (x). The order of calculationsremains the same.

Example 1. Given a differential equation y' =y-x \vith the initial con-dition y (0)= 1.5. Calculate to two decinlal places the value of the solutionof this equation when the ar~ument is x -= 1.5. Carry out the calculationsby a combined Runge-Klltta and Milne method.

Solution. We choose the initial Interval It frOll) the condl tion hoi < 0.01.To avoid involved writing, let us take h-=0.25. Then the entire interval ofintegration frool x=O to x= 1.5 is divided into six equal parts of length0.25 by means of points xi (i == 0, 1, 2, 3, 4, 5, 6); we denote by Yi and y;the corresponding values of the solution y and the derivative y'.

We calculate the first three values of y (not counting the initial one) bythe Runge-Kutta ruethod [frotH fonnulas (3)]; the remaining three values- Y4' Ys, Y6 - we calculate by the Milne I11ethod [from formulas (5)]

The value of Ye will obviously be the answer to the problem.We carry out the calculations with two reserve decilnals according to a

definite schenle consisting of h\'o sequential Tables I and 2. At the end ofTable 2 we obtain the answer.

Calculating the valueYl' Here, f(x, y)=-X+IJ, xo==-O, Uo=1.5

h = 0.25. !J.Yo = ~ (k~O) + 2k~O) -I- 2k~O) + k~O =

= ~ (0.3750+20.3906+20.3926+0.4106) =0.3920;

k~O) == f (xo, Yo) h == (- 0 -1- 1.5000)0.25 -= 0.3730;

(

k(O)

k~O)=f xo+~. Yo+-+) h=(-0.125+ 1.5000+-0.1875) 0.25= 0.3906;

(h k(O) )

k~O)=f xo+ 2 ' Yo++ h=(-O 125+1.5000+0.1953)0.25=0.3926;k~O) == f (xo-t-/z, Yo+ k~O h = (- 0.25 + 1.5000 +0.3926) 0.25 = 0.4106;

YI =Yo+ ~Yo== 1.5000+0.3920== 1.8920 (the first three decilnals In thtsapproximate number are guaranteed).

Let us check:

Ik~O) _k~O) I 10.3906-0.3926 1 20

8= k~O)_k~o) = 10.3750-0.39061 "'='156=0.13.

By this criterion, the interval h that we chose was rather rough.Sim ilarly we calculate the values Y2 and Y. The results are tabulated

in Table 1.

13*

388 ApproxImate Calculations

Table 1. Calculating .vI' YJ' .vI by the Runge-Kutta Method.f(x, y)=-x+y; h=0.25

[Ch. 10

I

, t(Xi+: 'Value of i xl YI Yi~ k(i) kIll)

k(i)

== f (xi, Yi) J,

Y'+T

0 0 1.5000 1.5000 0.3750 1.5625 0.39061 0.25 1.8920 1 .6420 0.4105 1.7223 0.43062 0.50 2.3243 1.8243 0.4561 1.9273 0.48183 o 75 2.8084 2.0584 0.5146 2.1907 0.5477

Value of it( Xi+i '

k(i) f (xi +h, k(i) ~Yik~i) ) a Yt-j- k~i 4

Yl+1

Yi+T

0 1 .5703 0.3926 1.6426 0.4106 0.3920 1 .89-20] 1.7323 0.4331 1.8251 0.4562 0.4323 2.32432 1.9402 0.4850 2.0593 0.5148 0.4841 2.80843 2.2073 0.5518 2.3602 0.5900 0.5506 3.3590

Calculating the value of .P4. We have: f (x, y) =- x+ y, h=O.25, x4 = 1;I Yo= 1.5000, YI = 1.8920, Y2 = 2.3243, Y. = 2.8084;y~ = 1.5000, Y; = 1.6420, y~ = 1.8243, y; = 2.0584.

Applying formulas (5), we find

- 4h "Y4=Yo +3 (2Y1-Ya+ 2y.) =

= 1.5000+4.0325 (21.6420-1.8243+2.2.0584) = 3.3588;if,= f (x4 , it) = - 1+3.3588 = 2.3588;== h -/ " 0.25Y'=Y2 +"3 (Y4 +4y. +YI ) = 2.3243+ 3 (2.3588+4.2.0584 + 1.8243)=3.3590;

I~-Y: 1_,3.3588-3.3590 , _ 0.0002 7 10-' < 1 0 001.s, 29 - 29 - 29:::::' 1"'.'

hence, there is no need to reconsider the Interval of calculations.

Sec. 5] Numerical Integration of Ordinary DiOerential Equations 389

We obtain Y4 =~= 3.3590 (in this approximate nu'mber the first threedecimals are guaranteed).

Similarly we calculate the values of y, and Yo' The results are given inTable 2.

Thus, we finally have

y (1.5) = 4.74.

4. Adams' method. To solve (1) by the Adams method on the basis ofthe initial data y (xo) = Yo we in some way find the following three valuesof the desired function y (x):

(these three values may b~ obtained. for instance. by expanding y (x) in apower senes (Ch IX, Sec. 16). or they may be found by the method of successive approximation (1). or by applying the Runge-Kutta nlcthod (2)and so forth].

With the help of the nurnbers xO' x.. x2 Xa and Yo. Yl' Y2. Ya we calcu-late qo' qt. q2' qat where

qo = IzY~ -= hf (xo Yo). qt = hy; = hf (XI' YI)'q2 = h Y~ === hf (XI' Y2)' qa = hy::= hI (xa Yale

We thrn form a diagonal table of the finite differences of q:

li.y= 1/ =f (x, y) q=y'h aq=Qn+l-q", li.2q= li.1q=-

x " ;; l/n+ I -lIn =li.qU+I-li.q,l =li.2q,.+ .-

-li.'Jq"

Xo I110 I ~!lo f (XOt Yo) I qo ~qo t:tlqo ~3ql)XI IY\ I ~YI f (XI' YI) I q, ~ql /).2QI ,\'q.x2 1 Y2! ~Y2 f (x 2 Y2) I q" ~q2 ~2q2 ~3q2XI II/I I ~Ya f (Xa, Ya) I qa ~qa &2qax4 1 Y41 ~Y4 t (X4t Y4) I q4 tiq4XII Y. I ~y, f (XI' UI) I q.XII Y.I I

390 Approximate Calcu lat ions [Ch. 10)

(7)

The Adam5 method consists in continuing the diagonal table of differen-ces with the aid of the Adanzs formula

A_I A 5 A2 3 AIu y,,--qn+2 uqn_l +12 u qn-2+"8 u qn-I'

Thus, utilizing the numbers qa, 6,q2' t12qh &aqo situated diagonally inIhe difference table, we calculate, by means of formula (7) and putting n::.:.- 3

in It, L\YI=ql+-} L\ql+~ L\lql + ~ L\'qo. After finding L\YI. we calculateY4= Ya + t1Ya And when we know X4 and U,a, we calculate q4= hf (x4, Y4)'introduce Y4' &Ya and q4 into the difference table and then fill into it thefinite differences 6,qa' 6,"q2' !J.3q1J which are situated (together with q4) alonga new diagonal parallel to the fIrst one.

Then, utilizing the nunlbers of the new diagonal, we use formula (8)(putting n=4 in it) to calculate ~Y4' Y5 and q, and obtain the next diagonal:qa, t1q., ~Iqa, 6,aq2' Using this diagonal we calculate the value of y, of thedesired solution y (x), and so forth.

The Adams formula (7) for calculating fiy proceeds from the assumptionthat the third finite differences 6,3q are constant. Accordingly, the quanti tv hof the initial interval of calculations is determined frol11 the inequalityh4 < lo-m [if we wish to obtain the value of y (x) to an accuracy oflO-m].

In this sense the AdalTIS formula (7) is equivalent to the fonnulas ofMilne (5) and Runge-Kutta (3).

Evaluation of the error for the Adams nlethod is complicated and forpractical purposes is useless, since in the general case it yields results withconsiderable excess. In actual practice, we follo\v the course of the thirdfinite differences, choosing the interval h so stnall that the adjacent diffe-rences ~3qi and 6,3 qi + 1 differ by not more than one or t\\70 units of the ~ivendecil11al place (not counting reserve desimals).

To increase the accuracy of the result, Adams' fOrlnula may be extendedby terms containin~ fourth and higher differences of q, in which case thereis an increase in the number of first values of the function y that are neededwhen we first fill in the table. We shall not here give the Adams fonnulafor higher aceuracy.

Example 2. Using the combined Runge-Kutta and Adanls nlethod, calculate to two decinlal places (when x = 1.5) the value of the solution of thedifferential equation y' = y-x wi th the Initial condi tion y (0) == 1.5 (seeExanl pIe I).

Solution. We use the values Yl' Y2' 1JI that we obtained in the solutionof Example 1. Their calculation i~ given in Table 1.

We calculate the suhsequent values Y4' Ys, y, by the Adams method (seeTables 3 and 4).

The answer to the problerTI is U, = 4.74.For solving system (4), the Adams formula (7) and the calculation scheme

shown in Table 3 are applied separately for both functions Y (x) and z (x).

Find three successive approximations to the solutions of thedifferential equations and systems indicated below.

3176. y'=XI_t-yl; y(O)=O.3177. y'=x+y+z, z'=y-z; y(O)=l, z(O)=-2.3178. y" = - y; Y (0) = 0, y' (0) = 1.

Table 2. Calculating J.'4' Y5' )'6 by the Milne Method.f (x, y) == - x +h; h = 0.25. (ItalIcised fIgures are Input data)

,--: i

Reconsider intervalValue U: =/ (Xi' Y.>

- = Y: =/ (X,. Y,) of calcula t 10 ns,ot I X, IJI. Y, IU.=! (x,. y,) y, 1. Yi following Indicationsof formula (6).

o I /1.500

1 1.5000 IIIJIIIIIIIIIII 1111111111111111111/11 111111l1li111/1 111111/111111/1111111111111/1111111111111/11111//111/1//1/11//1///11/1111/1111/111/11111

I 10, 25 11.8921 1.6420 1111111111l1li1111111111111111111111111111111111111111111/11111111111111//111111111111 I/JIIIIIIIIIIIIIIIII 111/11111111111[1111111111/11111111

2 /.512.32431 1.8243 /111111111111111111111111/111111111111111111111111111111111111111111111111111111111111111111111111111111 11/11111111111111/11111111111111111

i a /.75/2.80841 2.0584 11/1/1/111111111111111111[1111111111111111111111111 1111111111111111111 I 1111111111111111111111111111111/11111111111111111111111111111111111i - 4 /1.00 111111111111111111111111111111111111 3.3588 I 2.3588 I 3.3590 I ~71O-5 13.3590 I 2.3590 IDo not reconsider

5 11.251111111111111 111/1/1/111111111111111 3.9947 1 2. 7447 j 3.9950 j ~ 10-5 13.9950 I 2. 7450 I Do not reconsi,der

6 11.50 IIIIIIIIIIIIIIIIIIIIIIJIIIIIIIIIIIII 4.7402 I 3.2402 1 4.7406 j~ 1.4.10- 5 j~ll I Do not reconsider

Answer: I y (1.5) =4.74

Table 3. Basic Table for Calculatlni Y4' .vI' .ve by the Adams Method.f (x, y) = - x + y; h = 0.25

(Italicised figures are input data).....0

,qi= Y;h

Q.I t, Yi 6.Yl Y i= 6.q, li,2q, /i:lq,~ =f (xl, gil~

>-

01 0 I t .5000 111111111111111111 t .5000 I 0.3750 I 0 0355 I 0.0101 I0 0028"Ilo.25f t .8920 111111111111111111 1.6420 I 0.4105 I 0.0456 I 0.0129 I0 0037210 501 2.3243 111111111111111111 t .8243 I 0.4561 I 0.0585 I 0.0166 10.00473/0.751 2.8084 I 0.5504 I 2.0584 I 0.5146 I 0.0751 I 0.0213 I411 001 3.3588 I 0.6356 1 2.3588 I 0.5897 I 0.0964 I I511 251 3.9944 I 0.7450 I 2.7444 I 0.6861 I I I

Table 4 Auxi liary Table for Calculating by the Adams Method

t1y,=qi+ ; l\qi_l+ t21\2Qi -.+ ~ l\'qi-I

Value of I q, !.- 6.q t-1 .!.. 6. 2q'_2 .!. 6. 3q. dY,2 12

S &- 9

3 0.5146 0.0293 0.0054 0.0011 0.5504

4 o 5897 0.0376 0.0069 0.0014 0.63565 0.6861 0.0482 0.0089 0.0018 0.7450

ISec. 6] Approximating Fourier Coefficients 393

Putting the interval h = 0.2, use the Runge-Kutta method tocalculate approximately the solutions of the given differentialequations and systems for the indicated intervals:

3179. y'=y-x; y(0)=1.5 (O~x~I).

3180. y' = JL_ y2; y (I) = 1 (I ~ x~ 2).x3181. y'=z+l, z'=y-x, y(O)=l, z(O)=1 (O~x~l).Applying a combined Runge-Kutta and Milne method or

Runge-Kutta and Adams method, calculate to two decimal placesthe solutions to the differential equations and systems indicatedbelow for the indicated values of the argument;

3182. y' = x + y; y == 1 when x == O. Compute y when x = 0.5.3183. y' = x 2 -1- y; y = 1 when x = O. Compute y when x = 1.3184. y' =2y-3; Y-== 1 when x=O. Compute y when x=0.5.3185. {y' = -x +2y-+ z,

z' = x -t- 2y + 3z; y = 2, Z = - 2 when x = o.Conlpute y and z when x = 0.5.3186. {y' = - 3y-z,

z' =y-z; y=2, z=-I \vhen x=O.Compute y and z when x == 0.5.3187. y"=2-y; y=2, y'= -I when x=O.Cornpute y when x = I.3188. Y3Y"-i-1=0; y=l, y'=O when x=l.Compute y when x = 1.5.

3189. ~2t~+i-coS2t=0; x=O, x'-'l when t=O.

Find x (rt) and x' (n).

Sec. 6. Approximating Fourier Coefficients

Twelve-ordinate scheme. Let Yn =f (xn) (n = 0, 1, ... , 12) be the valuesof the function y = f (x) at equidistant poi nts xn = :It; of the interval lU,2n),and Yo - Yn We set up the tables:

IYO Yl Y2 Ya Y4 Ys Y,Yn YIO Y, Ya Y7Sums (~)Differences (l\) IUo U1 u2 Ua U. Us U,VI V2 Va V. Us

SumsDillerences

IUo Us U 2 uaUs Us U.SumsDifferences

394 Approximate Calculations [Ch 10

The Fourier coefficients an' bn (n = 0. I. 2. 3) of the function y = f (x)may be determined approximately from the formulas:

000 = 50 +S1+ 52 +5.. 6bl = 0.50'1 + 0.86602 +Ga.601 = to +0.866t l -l- 0.5t 2 6b2 = 0.866 ('t1+ 't2).6a 2 =50 -5.+0.5 (51-51)' 6ba=0'1- 0 a.6a l =to-t2 (1)

ya 1 1where 0.866 = -2- =:::: 1-10- 30

We havea

/(x) ~ a; +L (an cos nx +bn sin nx).n=J

Other schemes are also used. Calculations are simplified by the use ofpatterns.

Example. Find the Fourier polynomial for the function y= f (x) (0~x~2n)represented by the table

Yo

38

UI

38 12

Y.

4 14

Solution. We set up the tables:

y 138 38 12 4 14 4 - 1832 8 - 24 - 27 - 23

u 138 70 20 - 20 - 13 - 19 - 18v 6 4 28 41 27

I

38 70 20 - 20 I 6 4 28u _ 18 _ 19 _ 13 v 27 41

b I 20 51 7 - 20 0 I 33 45 28t 56 89 33 't -21-37

From formulas (1) we have

ao=- 9.7; a l = 24.9; a2 = 10. 3; a l = 3.8;bl =13.9; b2 =-8.4; b.=0.8.

Consequen tIy

f (x) ~ 4.8+ (24.9 cos x + 13.9 sin x) + (10.3 cos2x-8.4 sin 2x) ++ (3.8 cos 3x +0.8 sin 3x).

Using the 12-ordinate scheme, find the Fourier polynomialsfor the following functions defined in the interval (O,2n) by the

Sec. 6] Approximating Fourier Coefficients 395

tables of their values that correspond to the equid istant val uesof the argument., 3190. Yo=-7200 Ya==4300 Y6=7400 yg=7600

y, = 300 Yol == 0 Y" = - 2250 Y10 = 4500Y2 = 700 Ys = - 5200 Ys = 3850 Yll = 250

3191. Yo=O Ya=9.72 Ys=7.42 Ye =5.60YI -== 6.68 Yti = 8.97 Y7 == 6.81 y,o = 4.88Y2 == 9.68 Ys == 8.18 Ya= 6.22 YII = 3.67

3192. Yo-=::2.714 Ya=I.273 Ys=O.370 Ye =-0.357YI = 3.042 Y4 = 0.788 Y7 = 0.540 YIO = - 0.437Y2=2.134 Ys =0.495 Y8=O.191 YII-==0.767

3193. Using the 12-ordinate scheme, evaluate the first severalFourier coefficients for the follo\ving functions:

1a) f (x) = 2n 2 (Xl - 3nx2 -t- 2n2 x) (0 ~ x ~ 2n),

1b) f (x) = n 2 (x-n)2 (0 ~ x ~ 2n).

ANSWERS

Chapter I

1. Solution. Since a=(a-b)+b, then lal

Answers 397

V -e) x= I-ys (-ooO we determinethe points at which 1) y=O; 2) y=l; and 3) y=-1. When x~ -1-00,y --+ 1. 101. Hint. See Appendix VI, Fig. 14. 102. Hint. See Append1X VI,Fig. 15. 103. Hint. See Appendix VI, Fig. 17. 104. Hint. See Appendix VI,Fig. 17. 105. Hint. See Appendix VI, Fig. 18. 107. Hint. See Appendix VI,Fig. 18. 118. Hint. See Appendix VI, Fig. 12. 119. Hint. See Appendix VI,Fig. 12. 120. Hint. See Appendix VI, Fig. 13. 121. Hint. See AppendixVI, Fig. 13. 132. Hint. See Appendix VI, Fig. 30. 133.Hint. See Appendix VI.Fig. 32. 134. Hint. See Appendix VI, Fig. 31. 138. Hint. See Appendix VI.Fig. 33. 139. Hint. See Appendix VI, Fig. 28. 140. Hint. See Appendix VI.Fig. 25. 141. Hint.

398

Fornl a table of values=

o

x 0

y 0

Answers

2 3 I -1 -2 I -38 27 .. I -1 -8 1-274 9 .. I 4 I 9

Constructing the points (x, y) obtained, we get the desired curve (see Appen-dix VI, Fig. 7). (Here, the parameter t cannot be laid off geometrically!)142. See Appendix VI, Fig. 19. 143. See Appendix VI, Fig. 27. 144. SeeAppendix VI, Fig. 29. 145. See Appendix VI, Fig. 22 150. See AppendIx VI,Fig. 28. 151. Hint. Solving the equation for y, we get y= Y25 - x!. It isnow easy to construct the desired curve froln the points. 153. See Appen-dix VI, Fig. 21. 156. See Appendix VI, Fig. 27. I t is sufficient to construct

the points (x, y) corresponding to the abscissas x==O, i, a. 157. Hint.Solving the equation for x, we have x=..:. 10 log y-y 0) and calculating the abscissa x froln the formula SlIl :8' a) n~4; b) n> 10; c) n~32. 167. n >~e

-1 =N. a) N= 9; b) N=99; c) N=999. 168. ~=5 (8 < 1). a) 0 02;b) 0 002; c) 0.0002. 169. a) log x < -N when 0 < x < ~ (N); b) 2x > N when

7x> X (N); c) \ f (x) , > N when I x \ > X (N). 170. a) 0; b) 1; c) 2; d) 30 .1 3 3 1

171. 2". 172. 1. 173. -2' 174. 1. 175. 3. 176. 1. 177. 4". 178. 3. Hint.1

Use the formula 12 +22 +... +n2 =6"' n (n -1-1) (2n + 1). 179. O. 180. O. 18t. 1.182. O. 183. 00. 184. O. 185. 72. 186. 2. 187. 2. 188. 00. 189. O. 190. 1. 191. O.

1 a-I 2 1192. 00. 193. -2. 194. 00. 195. 2" 196. 3a l 197. 3x. 198. -1. 199."2 .

Answers 399

4 1 1 3 1200. 3. 201. 3 202. "9' 203. - 56' 204. 12. 205. 2' 206. -3. 207. 1.tIl a 5 I

208. 2Vx 209. 3 V x2 210. -"3' 211. O. 212. 2"' 213. -2' 214. "2'

215. O. 216. a) ~ sin 2; b) O. 217. 3. 218. {. 219. ~ . 220. n. 221. ~.

222. cos a. 223. - sin a. 224. 11. 225. cos x. 226. - ;2' 227. a) 0; b) I.2 1 1 1 2 1

228. n' 229. 2' 230. O. 231. - Y3' 232. 2 (n _m2). 233.2" 234. l.2 2 1 1 1

235. 3 236. n' 237. -"4' 238. 1t. 239. 4"' 240. 1. 241. 1. 242. 4'.3

243. 0 244. n' 245. O. 246. e- 1 247. e2 248. e- 1 249. e-&.~

250. eX. 251. e. 252. a) 1. Solution. lim (cos x) x ::= lim [1-(I-cos x)) x =x-+o x-.o

r 1 ]_2~in2f---x- ..\1 2sin2 _= ~i: 0 ( 1- 2sin2 -i)x = ~i~ l(1 - 2 sln2 i) 2

(2~fn:!.!..)lim I

x-+o x=e

(2sin

2

; ) [(Sin i)2 X2] xSince Iitn --- =-2lim -- -. =-21 Iitn-=O, it follows

X'-~o x x--+o x 4x x -+0 42"

] 1that lim (cos x) x = eO == 1. b) ..r-

e' Solution. As in the preceding

x~o r

.!- 11m ( - 251:' f) (_ 2 Si112i)) II )X2 x-+o X S' l'case (see a , 1m (cos x = e . Ince 1m 2 =

X-.O x~o X

= _ 2 limrl(Sin ~ )2 X~2] = _ .:- . It follows that lim (cos X)i> =e ~

x->o X 4x 2 x-+o2

= ';'e' 253. In 2. 254. 10 log e. 255. 1. 256. 1. 257. -~. 2.')8. I. Hint.Put eX -1 =a, where a ~ O. 259. In a. Hint. Utilize the identity a=e1n a

260. In a Hint. Put l.. = a, where a-+-O (see Exalnple 2.59) 261. a-b.n1

262. I. 263. a) 1; b) "2 . 264. a) -1; b) 1. 265. a) -1; b) 1. 266. a) 1; b) O.267. a) 0; b) 1. 268. a) -1; b) 1. 269. a) -1; b) 1. 270. a) -00; b) +00.

400 Answer~

271. Solution. If x kn (k =0. 1, 2... ). then cos! x < 1 and Y=0;1

but if x=kn, then cos1x=1 and y=1. 272. y=x when O-c;x< I; Y=2n

when x==l; y=O when x> I 273. Y=lxl 274. Y=-T when x O. 275. y= 1 when O

Answers 401

tinuity of the second kind. 328. x= 0 is a removable discontinuity. 329. x= 1is a discontinuity of the first kind. 330. x = 3 is a discontinuity of the firstkind. 332. x= 1 is a discontinuity of the first kind. 333. The function iscontinuous. 334. a) x=O is a discontinuity of the first kind; b) the functionis continuous; c) x= k'Jt (k is integral) are discontinuities of the first kind.335. a) x=/tt (It is integral) are discontinuities of the first kind; b) x=k(k ~ 0 is integral) are points of discontinuity of the first kind. 337. No, sincethe function y = E (x) is discontinuous at x= 1. 338. 1.53. 339. Hint. Showthat when Xo is sufficiently large, we have P (-xo) P (xo) < o.

Chapter II

341. a) 3; b) 0.21; c) 2h+h2 342. a) 0.1; b) -3; c) Va+h-Va.344. a) 624; 1560; b) 0.01; 100; c) -1; 0.000011. 345. a) a~x; b) 3x2Ax +

3 ( A 2 a. 3 2 3 (2. 2x L\x+ (~X)I . 2x + L\x .+ x ox) + (L\x) , x + x L\x+ L\x), c) - Xl (x+ L\X)2' x2(x+ AX)2 'x+L\x- Yx; 1 _ ; 2x(2\x_ 1); 2x(2~~ -1);

d) Yx+L\x+Yx e) L\x

f) In X~i\X; Lin (1+ ~X). 346. a) -1; b) 0.1; c) -h; O. 347. 21.

348. 15 cm/sec. 349. 7.5. 350. f (x + L\x) - f (x) . 351. f' (x) = Jim f (x +L\x) - f (x) L\x \x~o L\x

352. a) L\q>; b) dq> = lim L\q>, where q> Is the angle of turn at time t.l\t dt \t-+o L\t

353. a) L\T ; b) dT = lim ~T, where T is the temperature at time t.L\t dt 6/-'+0 ~t

354. dQ ;;:::: lim ~Q, \vhere Q is the quantity of substance at time t.dt \t -'0 L\t

355. a) ~m; b) litn L\m 356. a) _.-!.-::::::: - 0.16; b) -2~1 ~-O 238;L\x 6x-+oL\x 6

c) - ~l =:::: -0.249; y~=! = - 0.25. 357. sec' x. Solution.y' = lim tan (x + ~x) - tan x = lim sin ~ = lim sin L\x X

~x -+0 L\x \x -+0 L\x cos x cos (x + ~x) \x-+o L\xX Urn 1 = _1_ = sec2 x. 358. a) 3x!; b) _ .!.; c) _1_;~x ....o cos x cos (x + ~x) cos2 X x' 2 t"

d) --=..!- . 359. ~ Solution. l' (8) = lim f (8 -t- L\x) -f (8) =sin2 x 12 I1x-.o ~x

V8-l-~x-V8 . 8+~x-8= lin1 == 11m _AX~o L\x Ax-+o ~x lV(8 + ~X)2+ V (8 + L\x) 8+ VB!]

= lim 1 = .!.. . 360. f' (0) = - 8, f' (1) ==: 0,~x-+o V(8+ L\x)I+2 V8+L\x +4 12

I' (2)=0. 361. x,=O, x2 =3. Hint. For the given function the equationI' (x) = f (x) has the form 3x2 = Xl. 362. 30m/sec. 363. 1, 2. 364. - 1.

-1365. f' (xo) = -! . 366. -1, 2, tan q> .= 3. Hint. Use the results of Exa Inp Ie 3

Xo V (L\X)2 1and Problem 365. 367. Solution. a) I' (O) == linl 11111 J _.::: OO~

~x-+o Ax ~x.-+0V Ax

402 Answers

427.

425.

405.

V-b) /,(1):: 11m 1+&x-l= lim 1 =00; c) f'- (2k2+ 1 n)=

Ax-+o &x Ax-.+-o V (Ax)e

Icos(2k+ 1n+~x)l .= lim 2 = lim I sin &x I =_ I; i' + (2k + 1 n) =Ax-+-o &x Ax~-o &x 2

= lim (sin&xl =1. 36k. 5xC -12x!+2. 369. -.!.-+2x-2xl 370. 2ax+b.Ax-+-+ 0 &x 3

371 15x! 372 tm- 1 b ( ) tTn+n-1 37 6ax5

374 n --. rna + m+n . 3 ..r .. -2" a r Q2+b2 x1 3 8 ~ 2 8 4b

375. 2x I -fix l -3x-. 376. 3"x l . Hint. y=XIX I =x l 377. 3x1 V x -2a be-ad -2x2 -6x+ 25 1-4x- ----:v-. 378. 379. 380.

3x V Xi (e+dx)! (x2-5x+ 5)2 Xl (2x-l)21 4-2

38t. Y ( y-)I . 382. 5 cos x-3 sin x. 383. --. 384.z 1 - l sin2 2x (sin x-cos X)2

385. tlsin t. 386. y'=O. 387. cotx-+. 388. arcsinx+ y x S111 X I-xl

389 t 390 8 x (7) x 92 x x-2 93 5x"-x5

xarc anx. . xe x-I- . 391. xe. 3 e -xa. 3 . -e-x-.394. eX (cos x-sin x). 395. xleX. 396. eX (arc sin x+ y_l__) .397. x (2Inx-l) .

1-x! In2 x2 2 In x 2 21nxl.

398. 3x Inx. 399. -+-2-2. 400. -110--.401. slnhx+xcoshx.x x x xn x

2xcoshx-x2 sinhx -t h2 404 -3(xlnx+sinhxcoshx)402. cosh2 x 403. an x.. x In2 x-sinh! x .

-2xl

40 1. I + 1 .6. '1r arc sin 1 x ..r arc sin x.1-~ rl-~ fl+~

407. x-yX2="farccoshx 408. 1+2xarctanhx 410. ~(ax+b)2x2 Yx2 -1 (l_x2)2 e c

x2-1 -x411. 12ab +18b2y. 412. 16x (3 + 2X2 )3. 413. --. 414. V .

(2x-l)8 l-x2

bx2 r Va2 l-tan2 x+tanol x415. V (a + bXI )2 416. - V x2 -1. 418. cos2 x419. - ~. . 420.2-15 cosl x sin x. 42t. -16cos~ . Hint. x=sin-It +

2 sin2 x cot x sina 2t

+cos-2t. 422. sin x 423. sin1x 424. ;COsx+2sinx .(1-3cosx)a cos4 x 2 15sin x-l0 cosx

2 cos x + 3 sin x 13 V S1l1 X co~ X 426. 2 V I-Xl 'V I + arc sin x

1 3 (arc sin x)! - 1Y 428.2 (1 -t- x2) arc tan x y I-x! (1 + x2) (arc tan X)2

Answers 403

Solution.438.

a-3x

2 Ya-x 1+2 Y y

468.

x cos 2x2 sin 3xl

(x+ 2)'

470.

437.

2ex - 2x In 2 5 In4 x .V +--. 432. (2x-5) X3 (2eX - 2x + 1)2 x

433. -a sin (ax + ~). 434. sin (2t +q.

430.

436._2COS xsin' x

49t.

6Yy V(y+ y)2 473 1 474 a 2. ' ..r S1n X COS X.

f eX + 1477. x cos Xl. 478. 3t 2 sin 2t'.

4 cos 2x _(U-~) sin 2x . 483. O.479. 3 cos x cos 2x. 480. tan x. 481. -- 482. ~..rr======:::;;:===sin'" x 2 f a sin2 x + ~ cos2 X

1 arc sin x (2 Jrc cos x-arc sin x) 2 1484. - ..r 485. ..r . 486. --

2 , I-xi x , 2x2 -1 1+xl

487. xarccosx- yf=X2 .488. 1 .489." ("~ .490. 2 a 2 -x.(1-x2) /2 Ya - bxl V a+x-x. 492. arc sin yx. 493. .r 5

2x -Xl , 1-2Sxl arc sin 5x

469.

466.

464.

sin2 .:..a

1 (2x)' = 2 . 439. - 2 --1 - 1..r ..r ..r 440. ..r 441. --.f 1-(2x)2 f 1-4x2 X , ~:'-l 2 f x-x2 1+x2

442. 1-1 f . 443. _10xe- x2 444. _2tS- X2 In 5. 445. 2xl02x (1 +x In 10).+x-ex 2

446. sin 2t -t- 2f t cos 2t In 2. 447. ..r . 448. -- 449. cot x log e.f 1- e2X 2x 1- 7

-2x 21n x 1 (eX +S cos x) 1-x2-4450. 451. ----- 452. -_--.J

l-x2 X x In x (eX +5 sin x-4arc sin x) Y l-x21 1 1)

453. + 454. ..r-- + ..r- (1 +1n2 x) X (1 + x2 ) arc tan x 2x fIn x + I 2 (f x +X )

455. Solution. y' = (sin! 5x)' cos ; +sin! 5x (cos! ~ )' = 3 sin!5x cos 5x 5 cos! ~ +-I- sin! 5x 2 cos ; ( - sin i ){ = 15 sin! 5x cos 5x cos! i - fsin! 5x cos i sin i .456. 4x+3 457. x

2t4x-6 . 458. x

7.459. x-I .

(x- 2)8 (x- 3)5 (1_x 2 )5 x2 V 2t2 - 2x -t- 11 x2 (1 + Y X)3 5 V--a-2

460. -:;r--. 461. ..r 462. V 463. x (1 +x) f (a 2 -I ~ 2)3 f (1 + x2)' x

--:t-r====:::::l:::;::====::::::::.:::=:: . 465.V(x-l)3 (x --1- 2)5

2ubrnnx"- 1 (a 1- bA'l )m-l467.

(a_bx'l )Tn+J3x2 + 2 (a -1- b +c) x +ab + be + ac

2 Y(x+a) (x+b) (x+c)

471. 2 (7t .-!- 4) Vat + 2. 472. ..r y-af t2ay-y2)'

475. 1 476. 10 tan 5x sec! Sx.sin4 x cos4 x .

435.

eX +xex + 1429.

2 Vxex +xa

X cos(x'-5x+1)----X2 COS2 ~

x-1

404 Answers

Hint.

529.

53t.

494.1 ~na 1Y . 491. 498.

x l-ln1 x 1-2xcosa+xl 5+4sillx

497. 4x ' /6 x . 498. I ~nl ~ 499. a2

Y eax. 500. sin 2xe,fn2x V -x cos x501. 2m1p (2mamx +b)p-Iamx In a. 502. efJt (a cos pt-p sin Pt). 503. ea.x sin px.504. e- x COSI3x. 505. xn-1a- XI (n-2x2 In a). 506. - ~ y tan x (I + YCOs"'X In a).

3cot -In 3 y-507. ( \). 508. 2ax+b . 509. Y 1 810. V'

x sin x I ax2 +bx +c al +Xl I + xI -2 I x-I 2x+11

511. Y . 512. --. 513. -- tan -- . 514.2ax +x2 x Ina x x2 x x2 -x-2

3x2 - 16x + 19 Iy=5In(x-2)-3In(x+n.515. (x-I}(x-2)(x-3)' 516. sln'xcosx'

.r-- -6x 15a In2 (ax + b)517. r x2 - a

2 518. (3 _ 2xl ) 1n (3 _ 2x

')' 519. ax +b

520. Y 2 . 521. mx+n 522. Y2 sin In x. 523.x2 + a2 x2 -a2 sin' x

524. yf:t:Xi. 525. x + 1 . 526. Y 3 [ 2arc sin IX In 2+2 (1- arc cos 3x)}.X x l - I 1-9x2

527. (3:~: ~:-ln 3+slnllax ) a cos ax cos bX; b sin ax sin bx. 528. I 2' .cos bx cos bx + SUt x

I 530. Y 1 + In x -I- y 1x (1 + In2 x) 1- x 2 arc sin x x x 1-1n2 x

1 x! 2 x2 - 3x. 532. 533...F 534. --x{I+ln2 x) xC+x2 -2 cosx r sin x xoi_1

535 1 5 arc sin x 37 6' h2 2 h 2 ax ( 1 R -1+,. 36. 3/. 5 sIn xcos x. 538. e acoslpx+x (I _ t 2) /2

+ ~ sinh ~x)~ 539. 6 tanh 2 2x (I-tanh! 2x) 540. 2 coth 2x. 541. --,.-~'~-.l' aol +- Xl

I 1 -1 2542...r 543. -- . 544. - . 545. 546. x arc tanh xx r 1n2 x-I cos2x sin x l-x2547. x arc sinh x. 548. a) y' = I when x> 0; y' = - I when x < 0; y' (0) does

t exist; b) , (2 I 49 ' I 550 f'() { -I when x~ 0.no y -= x. 5 Y =x . x = _e- x when x > O.I ya ", 2

552. "2+3' 553. 611.554. a) f- {O)=-I, f+ (0)=1; b) f- (0) =(i ,, -2, , " ,

f + (0) = a; c) f _ (0) = I, f+ (0) = 0; d) f _ (0) = f+ (0) = 0, e) f_ (0) and. x-3f + (0) do not exist. 555. I-x. 556. 2+ -4-' 557. -I. 558.0 561. Solu-

tion. We have y'=e- x (I-x). Since e-x=JL. it follows that y,=.JL (I-x)x x

or xy' =- y (I - x) 566. (I + 2x) (I +3x) + 2 (1 +x) (I +3x) +3 (x + 1) (1 +2x).567. _ (x +2) (5.\2 + 19x +20) 568. x2 -4x + 2

(x + I)" (x + 3)5 2 Yx(x-I) (x-2)'

Answers 40&

a

b

x+yx-y

616.

571.

3x2 +5 V---xr (x-2)9(x2 "':"7x+l)569. 3(x' +I) xl+I' 570. (x-I)(x-2)(x-3)V(x-I)I(X-3)c'

I~XI + X-.~4 If 572. xX (l + In x), 573. xx' +1(1 + 2 In x).3 (x-I) 2 (x +2) 8 (x +3) 2

Y'X-2..574. Vx 1 ~n x ,575. x I ( 1+ ~ In x). 576. xXX xX(~ + In x+ In'x ).577. rio X (Sl~ x + cos x Inx). 578. (cos x)&tn x (cos x In cos x - sin x tan x).

579. (1+~r[ln(I+;)+I~X]' 580. (arctanx)X x

x[tnaretanx+(l+xltaretanx]' 581. a) x~=3(I~X');, 2 ' 10 3 2 t -1 -2t

b) xy 2-cos x : c) xy = ---:f. 582. 2" t. 583. t + 1 . 584. 1_/2.1+5e!

t (2 - tI) 2 t +1585. 1-2tl ' 586. 3 Vi' 587. t (t l + I) 588. tan t, 589.

b t t ' ~ -1 when t < 0, 93 -2eat590. - - an . 591. - tan 3t. 592. Yx = 1 h t 0 5 a w en > .594. tan /. 596. 1. 597. 00. 599. No. 600. Yes, since the equality is an Iden-

2 b2x x2 x(3x+2y) yYtity. 601. -5 602. --2- 603. -"2. 604. - 2+2 .605. - -.

ay y x y x_Vu 607 2y2 _ l_y3 608 10606. x 3(x!-y2)+2xy-l+3xyl+4y' 10-3~(jsy

y cos! Y y 1- x2- y2 2 '609. -1. 610. 1 2 611. - I + 2+~ 612. (x+y). 613. y =-x cos y x x Y L

I I ~=eY-I=x+y-I' 614. f+e x 615. XYy

617. cy + xV~ . 618. x In y -y .!L.. 620. a) 0; b) -.!..; c) O. 622. 45;cx-YX2 +y2 ylnx-xx 2

2arc tan 2~ 63 26'. 623. 45. 624. arc tan - ~ 36 21'. 625. (0, 20); (1, 15);

e-1 ( 1 I)(-2, -12).626. (1, -3).627. y=x2-x+ 1. 628. k=lT 629. "8' -16

I631. y-5=0; x+2=0. 632. x-I =0; y=O. 633. a) y=2x; Y= -2- x;b) x-2y-l=0; 2x+y-2=0; c) 6x+2y-n=0; 2x-6y+3:rt=-=O;d) y=-=x-I; y=l-x; e) 2x+y-3=0; x-2y+l=0 for the point (I, I);2x-y+3=0; x+2y-l=O for the point (-1, 1). 634. 7x-IOy+6=O,

n 2 Y"2IOx+7y-34=0. 635. y=O; (n+4)x+(n-4)y--4-=0. 636. 5x+6y-

-13=0, 6x-5y+21 =0. 637. x+y-2=0. 638. At the point (1,0):I-x h . ty=2x-2; y=-2-; at the point (2,0): y= -x+2; y=x-2; at t e pOln

3-x(3, 0): y=2x-6; y= -2-. 639. 14x-13y+ 12=0; 13x+ 14y-41 =0..-

406 Answers

640. Hint. The equation of the tangent is 22.. + 2Y = 1. Hence, the tangentXo Yo

crosses the x-axis at the point A (2xo' 0) and the y-axis at B (0, 2yo)' Findingthe midpoint of AB, we get the point (xo' Yo)' 643. 40 36'. 644. The para-bolas are tangent at the point (0, 0) and intersect at an angle

arctan ~ ~88' at the point (1, I). 647. 8t=8,,=2; t=n=2Y2.

648. 1~2' 652. T=2asin~tan~; N=2asin~; 8t=2asin2~tan~:1 31

Sn=asint. 653. arctanT' 654. 2+2cp. 655. St==4312a; Sn=a;

t=2na YI +4312 ; n=a Yl +4312 ; tan ~l=2n. 656. St=a; Sn= -; ;(Po

t=Va2 +Q:; n=~Va2+Q~; tanll=-q>o' 657. 3ctn/sec: 0; -9cln/sec

658. 15 em/sec. 659. - %- mjsec. 660. The equation of the trajectory isy=x tan u-g v :sin 2a

--- x2 The range is The velocity,2v:cos2 a g

1/ V o sin a-gtV v:-2vogtsina+g2fl; the slope of the velocity vector is vocosaHint. To detennine the trajectory, elinlinate the parailleter t from the givensystcnl. The range is the abscissa of the point A (Fig. 17). The projections

of velocity on the axes are ~. and ~ The magm tude of the velocity is

V (~~r+ ('~r ; the velocity vector is directed along the tangent to thetrajectory 66l. Diminishes with the velocity 0.4 62. ({, {) .

663. The diagonal Increases at a rate of -- 3.8 cln/sec, the area, at a rateof 40 cm2 /sec 664. The surface area increases at a rate of 0 211 In 2 jsec,

the volume, at a rate 01 0.0511: mlJsec. 665. ~ em/sec 666. The mass of the rod

is 360 g, the density at M is 5x gjcrn, the density at A is 0, the densityat B is 60 gJcIn. 667. 56x6 +210x4 668. ex2 (4xl -1- 2). 669. 2 cos 2x670. 2(1-x

2

) 671. -x 672. 2 arc tanx+~ .3 (I +X2)1 Y(al +x2)3 I -j-x2

2 2x arc slnx 674 1 t x 679 y'" -_ 6. 8 f'" (3 3'673. --+ 'i/" - COSal - ., 6 O. ) ==4 201- x2 ( I - x2 ) 2 a a

681. YV=(X~41)5' 682. yVI= -Msil12x 684.0; 1;2; 2. 685. The velocityis v=5; 4 997; 4.7. The acceleration, a=O; -0.006; -0.6. 686. The lawof motion of the point M 1 is x =a cos wt; the velocity at time t is-aw sin wt; the acceleration at time t is -aU>2 cos rot. Initial velocity, 0;initial acceleration: -am2 ; velocity when x=o is -aro; acceleration whenx=o is O. The maximum absolute value of velocity is au>; the InaximUlnabsolute value of acceleration is aw2. 687. y(n) =nlan. 688. a) n! (l_x)-(tz+l),

b) (_I)II+11

.:L

. ~~~3). 689. a) sin (x+ n ~ ; b) 2" cos (2X +n ~ ) ;2nx J

Answers 407

730.

c) (_3)n -'x. d) (_l)n-l (n-I)!. ) (-I)"+I.n!. f) 2n! .e , (1 +x)n' e (I +X)n-t 1 , (l-x)'Z+I'

g) 2n-I' ['2 +( -l)~J. h) (-l)n-J(n-l)!a" 690 ) x x51n x n 2' (ax + b)" . a x e +ne .\b) 2"-'e- zx 2(_I)"xZ+2n (-1)"-'x+ n (n2- 1) (-I )"-Z ]: c) (I-Xl) XX cos (X + n;) -2nx cos ( x+ (n-/) n) -n (n-I) cos ( x+ (n -;2) n) ;d) (-l)n-l.l.2... (2n-3) ('-(2 -I)]' ) (-I)n6 (n-4)! f 42n + I X n ,e n _ I or n~ _ x

2nx 2

691 y(h) (O):::-(n-I)! 692. a) 9t'; b) 2t 2 +2; c) -VI-t 2 693. a) -:-~t;aSlnb) 3 I. t . t: c) -I t ; d) t-:It' 694. a) 0; b) 2eat. 695. a) (I +tI) X

a cos sin 4a sin4 2" a sin

l+t -2e- t (d!Y)X (I +3t 2)., b) t 696 697 1(l _t)I (cos t + sin t)1 dx2 t=o =

699 2 cot t 700 4e2t

(2sin t - cos t) 701 _ 6 at (I + 3t + t2) 702 "t'" sint (sint+cost)s e.. m

d1x _ -1" (x). d1x 3 [f" (X)]2_f' (x)f"'(x) p2 b4703. dy2 - [I' (X)I ' d yl = [I' (X)]5 705. - it 706. - a2ys

2y! + 2 d2Y y d2xliiI 1707. -----ys-.708. dx2-=(I-y)l; dy2=Y2 709. 256 710. 16.

I 3a2 x71 t. a) 3~ b) - yS; . 712. L\y = 0.009001 ~ dy -== 0.009. 713. d (I-Xl) = I when

1x=1 and L\x=-a. 714. dS=2x!:J.x, L\S=2xL\X+(L\X)2. 717. For x=O.

n I718. No. 719. dY=-f2~ -0.0436. 720. dY=2700~0.00037.

1[ -mdx dx dx721. dY=-45 ~ 0.0698. 722. ---;;;+T 723. (I .)2 724. ..r--=

x -x r a2 -XZa dx -x2 -2dx I +cos cp725. --z-+2. 726. - 2xe dx. 727. In x dx. 728. -1-2. 729. - . 2 d

x

l'tdt IOx+8y -ye Ydx -Ldx. 734 ~+Ydx.-1+e2t. 732.- 7x+5y dx. 733. x x-y x-y

y2- xe y12 n

735. rrdx.737. a) 0.485; b) 0.965; c) 1.2; d) -0.045; e) 4+0.025~O 81.

738. 565 cnl'. 739. Y5~2.25; Y17:::::::4.13; Y70~8.38; Y640==:::253.740. VlO:::::=2.16; V70~4.13; V200:::::::5.85. 741. a) 5; b) 1.1; c) 0.93~

- (dX)2 -x (dX)2d) 0.9. 742. 1.0019. 743. 0.57. 744. 2.03. 748. a/ 749. '{ .

(I-xl) :I (l-x 2 ) 2

( I 2 cos x sin X) d 2 2ln X

s-3 (dx)!. 752. _ e-xx750. -Slnx nx+-x---,xr (x). 75t. x

2 '" 6 d)1 753 384 (dX)4 7 3 2n (2 5 + nn) (d 11X (x - ox+ )(x.. (2 _ x)s 54. sin x + 2 x) ..

408 Answers

755. excosa sin (x sin a +na) (dx)n. 757. No, since f' (2) does not exist.758. No. The point x= i is a discontinuity of the function. 762. 6=0.

14 n I763. (2, 4). 765. a) S=g; b) ;=4. 768.1nx=(x-I)-"2(x-I)2+

2 (x -1)1 h t: I +e( I) 0 e 1 769 Xl +x . t+ 3!;' ,w ere ~= X-, < I; a for n = 1;11:

1 Io for n < 1. 793. O. 795."5. 796. 12 797. -1. 799. 1. 800. ea. 801. 1.I I I

802 1 803. 804. -. 805. -. 806. 807. 1. 808. 1. 810. Hint.e e e

Find lim -i-, where S=~ (a-sin a) is the exact expression for the areau -~ 0 _ bh 2

3of the seglnent (R is the radius of the corresponding circle).

Chapter III811. (-00, -2), increases; (-2, 00), decreases. 812. (-00, 2), decrease~;(2, 00), increases. 813. (- 00, 00), increases. 814. (- 00, 0) and (2, 00),increases; (0, 2), decr~ases 815. (- 00, 2) and (2, 00), decreases. 816. (- 00, I).increas~s; (1, 00), decreases. 817. (- 00, -2), (-2, 8) and (8, 00), decreases.818. (0, 1), decreases; (1, (0), increases. 819. (-00, -I) and (1, 00), in

creases; (-1, 1), decreases 820. (- co, co), increases 821. (0, -}-). de-

.crea!>es; ( ~, (0). increases. 822. (-2, 0), increases. 823. (- 00 ,2), decreases;

Answers 409.

when x=l 844. Ymin == 1 wh~n

4846. Ymln=O when x-=O; !lm.lx ~e2

848. No extremunl. 849. Srl1al1esl1

value. M =."2 when x == 1. 850. nz = 01 n

when x=O nnd x= 10; M =5 for x=5. 851. m=2" when x=(2k +I) T;krt

M==l for x=2 (k=-=-O. 1. 2... ). 852. m-==O when x=l; A1=J1 \vhrnx=-I. 853. nl=-l \vhen x=-I; M=27 when x=3. 854. a) fn----6when x == I; M ~ 236 when x = 5; b) nt = -1579 when x = -10; M = 3745 \vhenx=12. 856. p=-2. q::..::.4. 861. Each of the terms must be equal to a

2

862. The rectangle must be a square with side f. 863. Isosceles. 864. Theside adjoining the wall must be twice the other side 865. The side of tllt:

cut-out square must be equal to i-. 866. The altitude must be half thebase. 867. That whose altitude is equal to the diameter of the base

868. Altitude of the cylinder, ~~; radius of its base R vi, where 1 Ymln=O when x=O; points of inflection, M 1 (; , 0) ;M. (arcsin ~2, 4~7); M.(n-arcsin ~2, _ 4~f). 966. EvenperiOdIC function with period 2~. On the interval [0, ~] Ymax = I \vhen

2 ( I) 2x==O; Ymax= lr- when x=arc cos - ,r- ; Ymin= - ,r- whenl 3r6 r6 3r 6

x=arc cos :6" ;Ymin = - 1 when x=n; points of inflection, M 1 ( ; ,0);M. (arc cos Y~. ~ y~); M.(arc cos ( - V~), -{ y~}967. Odd function. Points of inflection, M k (k~, k~) lk =0, I, 2, ... ).968. Even function. End-points, Al 2 (2 83, -1 57) Ymax == I 57 when x=O(cusp); points of inflection, M 1 ~ ( 1.54, -0.34). 969. Odd function.Litnlting points of graph (-I, '- 00) and (I, + 00). Point of inflection,0(0, 0); asymptotes, x= I. 970. Odd function. Ymax = ~ -I +?kn whenx = ~ +kn; Ymln= ~ n+ 1+2kn when X= ~ n+kn; points of inflection,

2k+1Mk(k~, 2k~); asymptotes, x= -2- 31 (k=O, I, 2, ... ). 971. Even

function. Ymin=O when x=O; asymptotes, fJ=- ~ x-I (as x -i""-oo) and

~y =2""x-1 (as x-++oo). 972. Ymin=O when x=O (node); asymptote, y=l.

973. Ymln = 1 + ~ when x= I; Ymax= 3: -I when x=-I; point ofinflection (centre of symmetry) (0, ~); asymptotes, y = x+2~ (left) and Y= x(right). 974. Odd function. Ymln=I.285 when %=1; Ymax=I.856 when

x=-I; point of inflection, M (0, ~); asymptotes, Y= ~ +n (whenx

x-.-oo) and Y=2 (as x-++oo). 975. Asymptotes, x=O and y=x-ln2.

414 Answers

976. Ymll1=1.32 when x=I; asymptote, x=O. 977. Periodic function with

period 2n. Ymln= i when x= i- n + 2kn; Ymax=e when x= ~ +2kn(

lr- }IS-I')r5-1 -(k=O, I, 2, ... ); points of inflection, M k ;lTCSin--2-+2k31, e 2

(Y5-1 YS+l\

and N k -arcsin-2-+(2k+l)n, e')' 978. End-points, A(O, I)and B (I, 4.81). Point of inflection, M (0.28, 1.74). 979. Points of innection~M (0.5, 1.59); asymptotes, y=0.21 (as x -+- 00) and y=4.81 (as x ~ -1- 00).980. The domain of definition of the function is the set of intervals (2k31,2k31 +31), where k = 0, 1, 2, ... Periodic function wit h period 231.

31Ymax=O when x=2"+ 2kn (k==O, I, 2, ... ); asymptotes, x=k31.

981. The domain of definition is the set of intervals [(2k - { ) n,

( 2k + -4- ) n], where k is an integer. Periodic function with period 2n.Points of inflection, M k (2kn, 0) (k ==0, I, 2, ... ); asynlptotes,

31X= T+2k31. 982. Domain of definition, x> 0; monotonic increasing

function; asymptote, x=O. 983. Domain of definition, Ix-2kn 1< ;(k == 0, I, 2, ... ). Periodic function with period 2rt Ynllll = I when

rtx==2kn (k==O, I, 2, ... ); asymptotes, x==T+kn. 984. Asymptote,

y = 1.57; Y -i""-1.57 as x -+ (linliting end-point). 985. End-points,1

AI, .(1.3I, 157); Ymln=O when x=O. 986. Ymln=( f)"e ::::::0.69 whenI

x == - ~ 0.37; y -+ I as x -+ + 987. Limiting end-point, A (+0, O)~eYmax = e e =::::: 1.44 when x == e~ 2.72; asymptote, Y.::= I; point of inflection,MI (0.58.0.12) and M2 t4 ~5, I 40). 988. xmin == -I \vhen t == 1 (y == 3); Ymtn = - twhen t = -I (x = 3) 989. To obtain the graph it is sufficient to vary t from to 2rt.xmin=- a when t = 3t (y == 0); xmax ==a when t = 0 (y = 0); Ymin -:: - a (cusp) when

331 ~.t=+T (x=O); Ymax==+a (cusp) when t==2 (x==O); pOints of inflection

31 3Jt 531 731 (a a )when 1=4' 4' T' T x= 2y"2" , Y= V2 .

I I990. xmin=-- when t=-I (y=-e); Ymax=- when t=1 (x=e); points ofe e

inflection when 1= - Y2, Le., (- :::~, - V 2eva) and when 1= V 2.

(- - )12'

I.e., Y 2 eV " eva); asymptotes, x=o and y=O. 991. Xmin= I and Ymln = 1when t=O (cusp); asymptote, y=2x when t -++OJ. 992. Ymin==O "'hen t=O.

Answers 415

X r ya4 _C2X2a y. 994 d d'993. ds=- dx, cos a=-; Sin a-= --. s=- 2 2 x;y a a a a -xa V aI_xi bx ..r--

~osa= ;sina-==- ,where c=-ra"-b2 995. ds=Va4 ---.c2x2 Ya2 - c2x!

1 y. p V'a =- Vp2+y2dx; cosa= ; sma= V . 996. ds= -dx,Y V p2+ y2 p2+ y" x

cosa== ~VXa; sina-==- VI. 997. ds=cosh~dx; cosa=_l_;a cosh ~

a

sin a'-" tanh : . 998. ds = 2a sin ~ dt; cos a =sin ~; sin a= cos ~ . 999. ds === 3a sin t cos t dt; cos a = -cos t; sin (J,:= sin t. 1000. ds == a V 1+ cp2 dq>; cos ~ =

1 a ..r--2 1 a= y . 1001.ds= 2 f 1+q> dcp; cos ~= - V-_. 1002. ds == -- dcp;

1-1- cp2 q> 1+

416 Answers

2xl Yx +1009. 5 x.4xm +n Yx 2x2n Yx2m + 2n + 1+ 4n + 1 1 x

Y7 arc tan 7".

1071.

4 5 I 7

/ - 9 - - 9 - - Xl1037. V nx. 1038. alx-Sa I X I +7a I X I -3.3x4 Vx 3xl Vx 63V- 2x2m yX

1040. --13- - 7 - x. 1041. 4m + 1..r- ..r- 2x'

1042. 2a r ax- 4ax + 4x r ax- 2xl + ../"_ 1043.5 r ax

1 \x- VIOl ..r- x1044. ...r- In V. 1045. In (x + r 4 +Xl). 1046. arc sin Y .2 f 10 x + 10 2 2

1047. arc Sin ~-In(x+yX1 +2). 1048*. a) tan x-x. Hint. Put tan1x=1

=sec2 x-l; b) x-tanhx. Hint. Put tanh2 x=1 hI. 1049.a)-cotx-cos x

-x; b) x- coth x. 1050. In(;e~ I' 1051. a In Ia C x I Solution. Sa a x dx ==-a5~(a-X)=-alnla-xl+alnc==alnl-C-I. 1052. x+lnI2x+ll

a-x a-x

Solution. Dividing the numerator by the denominator, we get ;;~~=

2 52x +3 5 52 dx Sd (2x + 1)=1+2x+l' Whence 2x+l dx = dx+ 2x+l=x+ 2x+l =3 11 x a

=x+ ln I2x+ll 1053. -2x+41nI3+2xl. 1054. 7]-bllnla+bxl.a ba-a~ x2 Xl

1055. aX+~lnlax+~I. 1056.2"+x+21nlx- 1 1. 1057. 2+2>:+x4 x'

-~ 1n I x +31 1058. 4"+3" +x2 +2x +31n Ix-II. 1059. a2x+2ab 1n! x-al-

b? 1 5 x dx S(x + 1) - 1x-a t060.~ In 1x+ll+ x + 1 Hint. (x+l)2= (x+l)2 dx=

Sdx \ dx ...r- 2 '1r ( a

= x+I-J (x+l)2' 1061. -2b f l-y. 1062. -3ij'" (a-bx).

t063. VXl +1. Solution.S.:dx = 21 Si X1 + I) = y' Xl + l. 1064. 2 yx+. x2 +1 Xl +1

lnlx 1 -./3 1 Ix Y7-2 Y21+-2 t065. ..r- arc tan x JI -5' 1066. .r- In Y Vf 15 4 ,,14 x 7 +2 2

1 Ir a+b+x Ya-b I ..r- x1067. II"" In - .. f"' Y . t068. x- r 2 arc tan 1r-'2 f a"-b2 r a+b-x a-b r 2

1069. - ( ~2 + ~I In Ia2-xll ) 1070. x- f In (x2 +4) +arctan ~ .1 ..r- ..~ 1 ... is

2 Y2 1n (2 r 2x+ r 7+8xl ). 1072. Vsarc sin x JI 7'

1 5 Ix ~r3 __ Y21 3 ... /S1073. 3"ln l 3x -21-2 f6 1n " V3+ Y2 1074. Y35 arc tan V 7"-

Answers 417

1131.

-{-In (5xl +7). 1075. i Y5x l + I + ;51n (xY5+ Y5x l + 1).107&. Yx l -4 +wr-- 1 I+ 31n I x+ r x2-41 . 1077. 2 In lxi - 5 1. 1078. 41n (2x 2 +3).

I 1 ax 1 Xl 11079. 2U In (a1x' + b2) +a arc tan b 1080. 2" arc sin (ii". 1081. 3" arC tan x'.

( arc tan -ir1082. } In Ix' + Y x6 - 1 I 1083. f Y(arc sm x)'. 1084. 4

1 (arc tan 2(,' r-------::::----1085. -8 In (1 + 4x2) 1086. 2 J! In (x+ YI + X2)~

3 2X

1087. - ~ e-"'''. 1088. -:31~4 42-'''. 1089. e1+e-1.l090. ; eli +2x-

a -~ 1 (aX bX) 2 - 2--2 ea. 1091. lib bX--x -2x. 1092. 3-1- Ya3X + W/ na~nan a In a r aX

1

1 X2 2 5 V-X1093. - 2

pX 2 +1. 1094. 2 In 7 7. 1095. -e x . 1096. li15

x 4

1097. Inle"-II. 1098. -,~ Y(a-be")' 1099. ~(ea +1)3. 1100. i-- 3 I~ 2 In (2" + 3) Hint. 2"~ 3 """ ~ ( 1- 2}:3)' 1101. Inla arctan (a").

1 11 +e- hX I 1t 102. - 2b In 1 _e- hX 1103. arc sin et 1104. - b cos (a +bx) ...r- x I .r-

'105. y 2 ~1l) Y"2 1106. x- 2a cos 2ax. 1107. 2 sin r x. 1108. -In lOXx Sitl 2~. . 1 x

xcos (log x) 1109. "2 --4- . Hint. Put Stll! x ="2 (I-cos 2~). 1110. 2 +S1n 2( . S . 09 1 co t ax

+-4- Hint. ee hint In 11 lIlt. -tan(ax+b). 1112. ----x.a attt3. alnltanil. 1114. ~lnltan(?;+i)l. ItIS. ~ Inltan axt b I.

I 2 I I I wr-1116. -2 tan (x). 1117. "2 cos (l-x ). 1118. x- V-2 cot x r 2-

_Y2Inltanx~I.I119. -lnlcosxl. 1120. Inlsinxl. 1121. (a-b)X

xlnlslna~bl. 1122.5Inlsin~l 1123. -2InlcosVXI.1I24. ~Inx I lit 1 6 a. I X 1 27 SI1)4 6xX Isin (x + I). 1125. n' an x I. 1 2 "2 sin a 1 ~.

1 1 I wr--1128. 4 4 1129. - -3 In (3 +cos 3x). 1130. - -2 f Ct S 2x.a sin ax

2 .r 32 3 t 4 x 1133 2 .r-t-,-- 9" r (1 +3 cos x) 11.. "4 an 3 ."3 r an x.5

1134. 3CO~ 'x. 1135.; (tan3x+cos\x). 1136. {(lnltan~I+2sll1ax)-

t4-1900

418 Answers

i- sinh (x'-I- 3).~2 arc tan ( t;:;) . Hint.

dx

==.\ta~~:2' 1188. j-Vlln(x-I-Y1-I-X'))" 1189.

1190 1 tanh % I Y2 "lr- -.3 3 1191. a) 1/;i arc cos - when x > y 2;n , ~ x

1187.

1 2 a . x 11137. 3Q1nlb-acot3xl. 1138. 5cosh5x-5s1nh5x.1139-2+4s1nh2~.

tt40. InItanh i I tt41. 2 arc tan e~. 1142. In 1tanh x I. tl43. In cosh x.5 1 I 1tt44. In I sinh x I 1145. - 12 V(5-x2)'. 1146. 4" In I~-4x + 1 . 1147.4 Y5 X

x. 1 -X2 ... 13 (ra)xarc tan Y5' 1148. -"2 e 1149. V 2" arc tan x 11 '2 -_ 1._ 1n (X Y3+ (2+3x2). 1150. Xa' - x2

1

+x-2In/x+1I1151. -}._.y 3 ,. eX1152. Inlx+cosxl. 1153. 3

1 (Inlsec3x+tan3xl+~3). 1154. --1-1-.sin x n x

1155. lnl tanx+ Ytanl x-21. 1156. Y2 arc tan (x Y2">- 4 (2X:+ I) .a SI n x V\Xa+ 1)2 1 1

1157. l"il""a. 1158. --2- 1159. 2 arcsin (x 2). 1160. a-tanax-x.

x sin x . tan x I ( x n ) I 3 V "1161. 2 --2-, 1162.arcsln-2-1163.alntan 2a+T .1164."4 (1+lnx).1165. -21n Icos Yx-II. 1166. {-In Itan i /. 1167. eare tan %+

1nl

(1 + x2) - I x I+ 4 +arctanx. 1168. -lnlsinx+cosxl 1169. Y21n tan 2 (2 -- 2x - Y2 cos ';'2' 1170. x+ ~21n / :~ ~~ /. 1171. In 1x 1+2 arc tan x.

sin 2X 5. X Y3 ..r--21172. e 1173. ya arcstn-2-+ y 4-3x. 1174. x-In(l+eX).

1175r 1 a;ctanx,la-+bb.1176.1n(eX+Ye2X-2).1177.-.!.lnltanaxl.f a2 -b2 V a a

T ( 2nt) 1 /2 + In xI (arc cos ~r1178. -2n cos r+q>o 1179. 4 1n 2-1nx 1180. 21181. - e- tan %. 1182. ~ arc sin (-7i) .1183. -2cot2x. 1184. (arc st X)2- Y 1-Xl. 1185. In (sec x -I- Y sec2 x -I- I). 1186. ~r- In I ~5 -I- 'lill 2x I.

4 ,. 5 f 5-sin2x

S dx S dx ----1+cos2 x sin! x + 2 cos2 Jt =-

Answers 419

c)8~(5r-3)'; d) ~ V(x+1)1-2VX+'1; e) In(sinx+Vl+sin'x).

1192. ~ [(2Xi;5)12 - 5 (2~lt5)tt]. 1193. 2(V;- ; +2 yX~2Inll+Vii) 1194. Inl ~2X+T-1 1. 1195. 2arctan Vex-I. 1196. Inx-ln2Inllnx+

y 2x+I+1

+21n21. 1197. (arc~tlx)l. 1198. }(tr-2) VeX + I. 1199. j.(COS'X-5)X

X yeGs x. 1200. In I 1 /' Hint. Put x=-tl . 1201. -i Vl-x'+I+Vx2 +1

1 x2 ..r-- 4 ..r-- '1r--+2 arc sinx. 1202. -'3 y2--x2 -:r y2-x2 1203. f x2 - a2 --a arc cos ~. 1204. arc cos 1.., if x > 0, and arc cos (- -!.) if x

420 Answers

Problem 12IS*. 1221. _ x c~ 2x +SillS2x. 1222. 2xl + I~X+ II sin 2x+

+2X: 5 cos 2x Hlnl. It is also advisable to apply the method of undetermined coefficients in the form

~ Pn (x) cos px dx = Qn (x) cos px + Rn (x) sin px,

where Pn (x) is the given polynomial of degree n, and Qn (x) and Rn (x) arepol ynomials of degree n with undetermined coefficients (see Problem 1218*).

Xl x' In x 11223. 3 In x- 9" . 1224. x In2x- 2x In x +2x. 1225. -- 2xl - - 4x2

- ..r- x2 + I X x2 11226. 2 V x In x-4 r x. 1227. -2- arc tan x-"2. 1228. 2 arc sin x-"4 X

X arcslnx+ ~ VI-xl. 1229. x In (x+ VI +x2)_ VI +xl 1230. -x cot x+1231. -~ + 1 It!-.I 1232. eX (sin x--cns x)+ In Isin x I sin x n an 2 2

1233 ax (sin x +cos x In 3) 1234 eax (a sin b"t-b cos bx) 1 35 !- [ . (I )_ 1+ (In 3)1 . a l + b2 2 2 sin n x

e- X2 y- (x 3-cos (1nx)). 1236. --2-(x1 +l). 1237. 2e x (Vx-l). 1238. 3-X2+)

Xl Xl x2-1 I-x In2x 21n x 2+3x Inx-g +2"-3x. 1239. -2- 1n I+x- x. 1240. --x---x--X-.

Xl xIII+ x21241. [In(lnx)-l).lnx. 1242. aarctan3x-T8+162In(9x2+1).1243.-2-X

1 ..r--x(arctanx)2-xarctanx+"21n(l+x2). 1244. x (arc sin X)2+ 2 r l-x2 X

. arc sin x I x I ..r--xarcslOx-2x.1245. +In V 1246. -2 y I-xxx 1+ I-xl

Xarc sin ..r x+2 ..r x. 1247. x tan 2 x+ In Icos 2x ,_ ~ 1248 e-x

f f 2 4 2' . 2 X

(COS2X-2Sin2x 1) 249 +xcos(2Inx)+2xSin(21nx)

X 5 1. 2 10 xl. x dx

1250. - 2 (Xl + 1) +2 arc fan x. Solution. Putting u = x and dv = (x 2 -t- 1)2 '1 S x2 dx x

we get du=dx and v= 2(x2 +1)' Whence (x2+1)2 2(x2 +1) +

Sdx xII (1 x+ 2(xl+l) 2(xl+l)+2arctanx+C. 1251. 2a2 iarctan(i+

+xl~al)' Hint. Utilize the identity 1.... ;1 [(xl+al)-xl ). 1252. 1- X..r-- at x ..r--

X f al -x'+2 arc sin tJ' Solution. Put U= f ai-xI and dv=dx; whence

dU=-VXdX andv=x;wehave fYal Xldx=XVa2-xl-S,::::~2dX=aI-xi II Va2-x2

., ygl_XI_s(al-xl>-alctx;::x Vaa-x,I-S val-xl dx+a2S dx .Yal-xi Val-x'

Answers 421

Consequently, 2 SVa2-x2dx=x a2_-x2+a2arcsin : 1253.i VA +x2++ ~ In I x+ VA +x" I. HInt. See Problem 1252*. 1254. -i V9-x2++ ~ arc sin i. Hint. See Problem 1252*. 1255. i- arc tan x; I. 1256. ~ X

I x I 2 6x-1 1 ~ 7X In x+ 2 . 1257. Vll arctan VIT' 1258. 2 In (r-7x+ I. + V3 X2x-7 3 5

xarc tan V3 1259. 2 1n (x2-4x+5)+4 arctan (x-2). 1260. x-2'ln (x2+

9 2x+3+3x + 4) + y"7 arc tan "7 . 1261. x + 3 In (x l -6x + 10) + 8 arc tan (x -3).

1262. /-2 arc sin~ 1263. arcsin(2x-I). 1264. InIx+ f + VXl + px+q I.1265. 3 Vx2-4x+5. 1266. -2 Vl-x-x2 -9arc sin W.1267. ~ V5xl -2x+ I + 5 ~'5 In ( x V5- ~5 + V5x l -2x+ I)

1268. In I V \. 1269.-arcslD 2-:-:,. 1270. arcsin 2--V- (x> V2).1+ l-x2 Xy 5 (I-x) 2

1 x+l.r Ir,1271. -arCS1tl

x+

I' 1272. ~ y x~+2x+5+2In(x+l+, x2 +2x+5).

2x-1 .r-- I . 2x+ 1 .r1273. -4- f x-x2+'8arcS10(2x-l)o 1274. -4- y 2-x-x2+

9 2x+l IIXl - 3 1 1 3-~1flX+- arc S111 -- 1275. -4 In -l-I . 1276. - ..r- arc tan V- 8 \i x- y 3 J

1277. In ( r +{ + VI +eX +e2X ) 1278. -Inl cosx + 2+ Vcosl x+ 4CI,SX + 11.

1279. - V'1-4 In x-Inl x- 2 arc sin 2 -:;~ x . 1280. ~b In Ix +-1- bI., 5 0- x a

1281. x+3In/x-31-3In)x-21 1282 ..!.-lnl(X-l)\~+3)~112 ,X +2)4

1 161

I(x-l)4 (X-4)51 x 2 (X-4) 61283. In \x + :3)7 1284. 5x +In 2.(x-I) I

.1285I~X+ln IX~II.

1286. ~x+~lnl(2X_I/~~X+1)81 1287. ;-(X~2)i-X~2.___9__ I 1289. 8 _ 27 ~ In 1~-5 I

1288. 2(x-a) 2(x+I)O 49(x-5) 49 (X+2)+\i4J x+2

1290. -2(X2 _13X+2)2' 1291. %+Inl .,,-ml 1292. x++ln 1;::1-

-i arc tan x. 1293. Sl2 1n IX-31-2~ In Ix- 1 1+ 6~ In lx2 +4x +5) + I~O X

422 Answers

t 2 1294. 61 In .(x + 1)11+ .;_ arc tan ~r-l . 1295. .~_ Xxarcan(x+). x-x+ r3 r~ 4r2

x! + x Y2 + 1 Y2 x Y"2 1 x2 + X + 1 IX In xl-x JI- 2+ 1 +-4- arc tan I-xl' 1296. Tin xl-x+ 1+ 2 Y3 X

x'-I x arc tan x 2x-lxarctan XYa ' 1297. 2(1+x2)+ 2 1298. 2(x2+2x+2)+

x+2 5 2x+1+arctan(x+I). 1299. Inlx+II+3(X2+x+l)+3 Y3arctan ya-

1 3x-17 I 15- 2 In (x'+x+ 1).1300. 2 (x2- 4x +5) +2 1n (x2 -4x+5) + 2" arc tan(x-2).

_x2 + x I I I1301. 4 (x+ 1) (x2+ 1) +"2 ln Ix+ II-Tin (Xl + 1)+4 arctan x.

3 x 3 IX-II 15x5 +40xs +33x1302. -Sarctanx- 4 (X4 _1)+16 In x+l 1303. 48(I+x2)' +. 15 x-I+48arctanx. 1304. x x2_2x+2+21n(x2-2x+2)+3arctan(x-l).

1 1t305. 21 (81nl xl +81-1n I xl + 11) 1306. 2' ln lx4- 11-

I ~ I 12X4 + 1- Y51 13 3- -4 In I x +x4-II--;y=-ln Y' 1307. 2 ( 4)3 +-4 +

2 r 5 2x4 + I + 5 x- x-+2Inl;_~I. 1308. ~ (2InIXlx~II-~I-XI~I)' 1309. xii +

IX-2 1 1+In x-I. 1310. In)xl-7 In lx

7+11. Hint. Put 1=(x7 +1)-x7

. 1 5 1 1 I1311. In I x I-SIn) x + 1 1+ 5 (xs+ I)' 1312. 3" arc tan (x+ 1)-6 arc tanx

x+1 1 I I I I IX -2- 1313. g (x-I). 4 (x-I)! 7 (X-I)7' 1314. - 5x5 + 3x3 -X--

-arc tan x. 1315. 2Yx=I[(X-;I)"+3(XsI)2+ xl. 1316. l~alXX[2 V(ax+b)S-5b V(ax+ b)Zl 1317. 2 arc tan Y x+l. 1318. 6 V"X -

V - lr- h/-) 6 "/- 6"/;5 3 V--3 x+2r x-6In(l+ 11 x. 1319. "7 x V x- S V X5 _ 2 r++2Vx-3 Vx-6vx-31n II + Vx 1+ 6 arc tan vx.1320. InI(JrXT~_1)2 1- .;- arc tan 2 Y7 + I . 1321. 2 rX-2 V'2x

x+2+ x+ I r 3 3... /x lr- VX~-I 1

xarc tan r 2. 1322. -2 arc tan r I-x. 1323. 2 (x-2) +2 In I x +lr-~- 1 Z2+ Z + 1 2 22+ 1 2z

+, x -II 1324. 3 1n (Z_V+Y3arctan Y3 +zl-l' where

vx + 1 V2x+3 2.x+3 .r 1z= x-I 1325. ---x-' 1326. ---8- r x2 -x+l+ 16 In(2x-l+

Answers 423

21339. -cosx+3" coss x -

where

'If , 7 8+4x'+3x4 ..r-l-I 13 8 ( S 5 3 1 s)+2 f x -x+l). 132. - IS f -x. 2. 16 x-:24 x +"6 x X..r- 5 ..r- (1 3 ) ..r- 3 1X y 1+ x2 -16 1n (x+ y 1+ x2). 1329. 4x4 +8x! r x2 - 1-8 arc sln-X

1 ..r-z-- 1 1 2x-l ..r 2 191330. 2(x+l)2 r x +2x-"2arcSlnx+l 1331. -4- y x -x+l+S lnx

1 1-b- x' 1 Vx- 4 + 1+1X (2x-I+2 Yx'-x+I). 1332. -2 -;r-----==-. 1333. -4 In V---

y 1+2x2 ,. x- 4 +1-1

(2x! - 1) Y 1+ Xl 1 (z - 1)'1334. 3xs 1335. TO In z~+z+ 1 +

VlXi 1336. 1 4+3x3

Z == +. - 8" x (2 -t- xa) 2/,

- -} arc tan Vx- + 1.va 2z+ 1+-S- arc tan 73

V I1337. - 2 (x" + 1)1. 1338. sin x- ; sin' x.1 .. 1340 stn' x slnl x 1 x 1 X s,ft' X

1341" 1342. -'2'---Seas x. -3---S- -;rcos 2-:rcOS 21. 3x sin 2x sin 4x

2 sin' x - 21n I Sln x I 1343. T - -4-+~ x sin 4x x sin 4x sina 2x 5 1 1

1344. 8-32. 1345. 16-64+----:ta 1346. T6x+i2sin6x+64 sin 12x+1 cotl x 2 1+ 144 sina6x. 1347. -cot x - -3- . 1348. tan x + "3 tans x + "'5 tanS x.

cota x cotS x tanS x 11349. --3-- -S- 1350. tan x +-3- - 2 cot 2x. 1351. "2 tan' x+

3 1 1 I xl Y2-I 3 In I tan x 1- 2 tan2 x - 4 tan4 x 1352.~+21n tan 2 . 1353. -2- Xcos "2

X [In Itan i I+ In Itan ( i + ~ ) I] 1354. 4 s~::: - : ~:~I: +f In Itan i Isin 4x 3 sin 4x 3 I ( 1t ) I 1

1355. 16 cos4 4x +32 cos24x +32 1n tan 2x +""4 . 1356. 5" tan Sx -x.cot2 X 1 3 x

1357. - 2-- 1n I sin x I 1358. - 3" cot' x+cot x +x. 1359. 2 tan! 3" +x x I x I Xl sin 2\:1 coP x

+tanla-3tan"3+31n cos "3 +x. 1360. 4--8- 1361. --3-.

1362. - ~ V cos x + ~ V COSiO x-l~ V cos l ' x. 1363. 2 Vtan x. 1364. 2 ~2>

434 Answers

along the large lower side of the rectangle, and the y-axis, perpendicular toV 2

it in the middle. 1778. Solution. s=5~ dv; on the other hand, :~ = a,v.

t',

whence dt =1. do, and consequently, the acceleration time is t = rdv =5.a J a

tilx

1779. Mx=- S~ (x-t)dt+ ~ x=- ~ [xt- ~]:+~ x= ~x (1-+) .o

x

1780. Mx=-5(x-t) kt dt +Ax=~W-x2 ), 1781. Q =0.12 T RI: cal. Hint.o

Use the J6ule-Lenz law.

Chapter VI2 2 ~r1782. V=3(y2_ X2)X. 1783. S="3(x+y) r 4z2-l-3(x-y)2.

(1 ) 5 y2_x2 X2_ y2 y2_x2

1784. f -2' 3 =-3; f(l, -1)==-2. 1785. 2 ' 2 ' 2 'xy xy xy2xy R4 VI-I-x2

-2--2 1786. f(x, x2)=I+x-r. 1787. z=-IR2. 1788.f(x)=x -y - x

Hint. Represent the given function in tile form I ( ~ ) = V(;r+ I andreplace f by x. 1789. f(x, y)=x2 -; xy. Solution. Designate x+y=u,

u-t;v u-v u+v u-v (U-V)Ix-Y=v. Then x=-2-' Y=-2-; f(u, v)=-2- -2-+ -2- =, u2 -uv=-'-2- It remains to name the arguments u and v, x and y. 1790. f (u)=

=u2 +2u; z=x-I+ VY. Hint. In the identity x=l+f(VX"-I) putVx-I = u; then x =;= (u + 1)2 and, hence, f (u) = u2 + 2u. t 791. f (y) == VI +y2; z= VX2+y2 Solution. When x= I we have the identityy I +y2 =1f ( t ), i. e., f (y) = y 1+y2. Then f ( ~ ) = y 1+( ~ randZ=X V 1+ (~r = JI x"+y2. 1792. a) Single circle with centre at origin,including _ the circle (x2 +y2~ 1); b) bisector of quadrantal angle y = x; c) half-plane loeated above the straight line x+y=O (x+y > 0); d) strip containedbet~een the straight lines y= I, including these lines (-I ~y~I); e) asquare formed by the segments of the straight lines x= 1 and y= I, includ-ing its sides (-I ~x~ I, -1 0, x

Answers 435

par~bola Y = - x2 (x2 + Y > 0); .k) the entire xy-plane; I) the entire xy-plane,with the exception of the coordinate origin; m) that part of th~ plane locatedabove the parabola y2=X and to the right of the y-axis, including the pOtntsof the y-axis and excluding the points of the parabola (x~ 0, y > Yx);n) the entire place except points of the straight lines X= 1 ~nd y =0; 0) thefamily of concentric circles 21tk~x2+y2~n(2k+l) (k=O, 1, 2, ... ).1793. a) First octant (including boundary); b) First, Third, Sixth and Eighthoctants (excludIng the boundary); c) a cube bounded by the planes X:c:: I,y =~ 1 and z== 1, including its faces; d) a sphere of radius 1 with centreat the origin, including its surface 1794. a) a plane; the level lines arestraight lines parallel to the straight line x+y=O; b) a paraboloid of revo-lution; the level lines are concentric circles with centre at the origin;c) a hyperbolic paraboloid; the level lines are equilateral hyperbolas;d) second-order cone; the level lines are equilateral hyperbolas; e) a paraboliccylinder, the gcneratrices of which are parallel to the straight linex+y:+-I=O;the level lines are parallel lines; f) the lateral surface of a quadrangularpyramid; the level lines are the outlines of squares; g) level lines are parapolas y =- Cx2 ; h) the level lines are parabolas y = C Yx; I) the level Hnesare the circles C(t2 +y2)=-2x. 1795. a) Parabolasy=C-x2 (C>0); b) hyper-bolas xy = C (I C I~ I); c) circles x2+ y2 = C2; d) straight lines y = ax +C;c) straight lines y-=CX (x =1= 0). 1796. a) Planes parallel to the planex -t- Y -1- z --= 0; b) concentric sphere~ \vith centre at origin; c) for u > 0,one-sheet hyperboloids of revolution about the z-axis; for u < 0, two-sheethyperbololds of revolution about the same axis; both families of surfacesare divided by the cone X2+y2_ Z2=0 (u=O). 1797. a) 0; b) O;c) 2~d) e'l; e) linlit does not exist; f) linlit does not exist. Hint. In Item(b)pass to polar coordinates In ltcnls ~e) and ef), consider the variation of xand y along the straight lines y == kx and show that the given expressionmay tend to different lilnits, dependin~ on the choice of k. 1798. Continuous.1799. a) Discontinuity at x=O, y==O; b) all points of the straight linex==y (line of discontinuity); c) line of discontinuity IS the circleX2 +y2= 1; d) the tines of discontinuity are the coordinate axes.

1800 Hint. Puttin~ Y= YI == const, \ve ~et the function l, we ~et z == sin 2'Pt \vhence it iseviden t that if x ---+ 0 and y -+ 0 in such nlanner that q> = const (0~ q> ~ 231),then z -+ sin 2({). Since these linliting values of the function z depend on thedirection of cp, it follows that z does not have a lilnit as x -+ 0 and y -. o.

dz 2 az 2 az 2y az 2x1801. dx=-=3(x -ay), ay=3(y -ax). 1802. ax=(x+y)2' ay=-(X+y)2

1803. az=_ y az=~ 1804 az_ x az__ yax x 2 ' ay x ax- X2 _ y2' ay- Yx2_y2

1805. ~= y2 az xy 1806 dz _ 1 iJz _ax (x 2 -1_ y2)3!2 ' ay (x 2 +y2)3!:. . . ax - Y x2 + y2' ay-

y az y az x az= Yx2+ y2 (x + Yx2+y2) 1807. ox = - x2+y2' ay = x2+y2 1808. ax =yxY - I

436 Answers

iJz sin .1L sin JL..:3y=xY lnx. 1809. ~=_1!.., "'"cos 1L oZ::z.!..e xcoslL. 1810. ~~

u ox Xl X ay X X iJx

xy'l Y2x I - 2y OZ yxl 2x l - 2y az 1 X +a- JY I(xl_yl)' at!:::a - Iy I (Xl_yl)' 1811. ox=Yi cot yi'~%= _ X~ra_ cot x.~~. 1812. iJU -== yz (xy)Z-I. iJau ==xz(xy)Z-I. ~u= (xy)Ztn(xy).uy 2y , Y ,. Y X Y u z

iJu OU iJu _ ' 1'.1813. ox=yzxYlnz, iJy=xzxYlnz, oz==xyzxy I. 1814. fx (2,1)=2'

., , 1 z 1'1/(2, 1)=0. 1815. Ix (1,2,0)= 1, ' , (1.2.0)=2' f (I, 2,0)=2 .

x y x21820. - 3/ 1821. r. 1826. z=arc tan -+

Answers 437

1 az y. dz 1862 az . a.V-I. dz-slnx n sin x). 1861. a-=-~+2' d-=-l-,. a-=U~ , d-=x x Y x +x x x[

Y ] az, , az ,=xY q>'(x)lnx+-X .1863. ax=2xfu(u,v)+yrYfv(u, V); ay z:-2Yfu (U,v)+

+xrYf~(u.v). 1864 ::=0, ;;=1. 1865. :~=Y(1-:2)f'(XY+~);:; = (x+ ~ ) f' (xy + ~). 1867. :~=f~(X. y. z)+

438 Answers

a"z '" """ I " ,,, I"1905. ax" = fuu (q>x) "+2fuvyy+fv'l'yy

1914. u(x, y)= (y)+'I' (y). 1916. d2z=eXY XX [(y dx + X dy)" + 2dx dy]. 1917. d"u =